5 Marks Questions

Question 15 Marks

Prove that the radii of the circles x² + y² = 1, x² + y² - 2x − 6y - 6 = 0 and x² + y² - 4x - 12y - 9 = 0 are in A.P.

Answer

The given equation of circle are.

x² + y² = 1 .......... (1)

x² + y² + 2x + 6y - 6 = 0 ......... (2)

x² + y² - 4x - 12y - 9 = 0 ......... (3)

Respectively If a, b, care in AP, then b $\frac{\text{a}+\text{c}}{2}$

Let C₁ C₂ & C₃ are the centres of (1) (2) & (3)

For a = 1, b = 4, c = 7, $\frac{1+7}{2}=4\text{b,}$ therefore 1, 4, 7 or The centres of the three circles lie In AP.

$\therefore$ R₁= 1

R₂ = $\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{1^2+3^2+6}=\sqrt{16}=4$

R₃ = $\sqrt{\text{g}^2+\text{f}^2+\text{c}}=\sqrt{2^2+6^2+9}=\sqrt{49}=7$

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Question 25 Marks

Find the equations of the circles passing through two points on y-axis at distances 3 from the origin and having radius 5.

Answer

Let the required equation of the circle be (x - h)² + (y - k)² = a²
The circle passes through the points (0, 3) and (0, -3).
$\therefore$ (0 - h)² + (3 - k)² = a2 ........ (1)
And, (0 - h)² + (-3 - k)² =a² ....... (2)
Solving (1) and (2), we get:
k = 0
Given:
Radius = 5
$\therefore$ a² = 25
So, from equation (2), we have:
h² + 9 = 25 ⇒ h = ±4
Hence, the required equation is (x ± 4)² + y² = 25, which can be rewritten as
x² ± 8x + y² − 9 = 0

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Question 35 Marks

The circle x² + y² - 2x - 2y + 1 = 0 is rolled along the positive direction of x-axis and makes one complete roll. Find its equation in new-position.

Answer

Given circle is x² + y² - 2x - 2y + 1 = 0

Rewriting the equation, we get,

x² - 2x + 1 + y² - 2y + 1 = 1

(x - 1)² +(y - 1)² = 1 ........ (1)

The given circle has its centre at (1, 1) and radius = 1 from (1).

When circle is rolled on x-axis, it center moves horizontally through distance Figure shows circle with centre (1, 1) at P.

After rolling it on X-axis, it takes the position Q.

The coordinates of it's centre become $=(1,\ 1+2\pi)$

Radius of the circle at Q = 1,

Hence, equation of new circle is.

$[\text{x}-(1+2\pi)^2]+(\text{y}-1)^2$

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Question 45 Marks

Prove that the radii of the circles x² + y² = 1, x² + y² - 2x − 6y - 6 = 0 and x² + y² - 4x - 12y - 9 = 0 are in A.P.

Answer

The given equation of circle are.

x² + y² = 1 .......... (1)

x² + y² + 2x + 6y - 6 = 0 ......... (2)

x² + y² - 4x - 12y - 9 = 0 ......... (3)

Respectively If a, b, care in AP, then b $\frac{\text{a}+\text{c}}{2}$

Let C₁ C₂ & C₃ are the centres of (1) (2) & (3)

For a = 1, b = 4, c = 7, $\frac{1+7}{2}=4\text{b,}$ therefore 1, 4, 7 or The centres of the three circles lie In AP.

$\therefore$ R₁= 1

R₂ = $\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{1^2+3^2+6}=\sqrt{16}=4$

R₃ = $\sqrt{\text{g}^2+\text{f}^2+\text{c}}=\sqrt{2^2+6^2+9}=\sqrt{49}=7$

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Question 55 Marks

The circle x² + y² - 2x - 2y + 1 = 0 is rolled along the positive direction of x-axis and makes one complete roll. Find its equation in new-position.

Answer

Given circle is x² + y² - 2x - 2y + 1 = 0

Rewriting the equation, we get,

x² - 2x + 1 + y² - 2y + 1 = 1

(x - 1)² +(y - 1)² = 1 ........ (1)

The given circle has its centre at (1, 1) and radius = 1 from (1).

When circle is rolled on x-axis, it center moves horizontally through distance Figure shows circle with centre (1, 1) at P.

After rolling it on X-axis, it takes the position Q.

The coordinates of it's centre become $=(1,\ 1+2\pi)$

Radius of the circle at Q = 1,

Hence, equation of new circle is.

$[\text{x}-(1+2\pi)^2]+(\text{y}-1)^2$

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Question 65 Marks

Find the equation of the circle which passes through the points (2, 3) and (4,5) and the centre lies on the straight line y - 4x + 3 = 0

Answer

Find the equation of the circle which passes through the point (2, 3) and ( 4, 5)
and the centre lies on the straight line y - 4x + 3 = 0.
Let the equation of required circle be x² + y² + 2gx+ 2fy + c = 0 which passes
through the point (2, 3) and ( 4, 5) .
$\therefore$ 13 + 4g + 6f + c = 0 ......... (1)
41 + 8g +10f + c = 0 ........... (2)
Centre (-g, -f) lies on y - 4x + 3 = 0
-f + 4g = -3 .......... (3)
Subtracting (1) from (2), we get
28 + 4g + 4f = 0 ........ (4)
Solving (3) and (4) we get,
f = -5 and g = -2
Substituting values off and g in (2) we get,
41 - 16 - 50 + c = 0
c = 25
$\therefore$ The required equation of the circle is, x² + y² - 4x - 10y + 25 = 0.

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Question 75 Marks

Show that the point $(\text{x},\ \text{y})$ given by $\text{x}=\frac{2\text{at}}{1+\text{t}^2}$ and $\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}^2}\Big)$2 lies on a circle for all real values of t such that $-1\leq\text{t}\leq1,$ where a is any given real number.

Answer

$\text{x}=\frac{2\text{at}}{1+\text{t}^2},\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}}\Big)$
$\text{x}^2+\text{y}^2=\frac{4\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}+\frac{\text{a}^2(1-\text{t}^2)^2}{(1+\text{t}^2)^2}$
$=\frac{4\text{a}^2\text{t}^2+\text{a}^2(1-2\text{t}^2+\text{a}^2)\text{t}^4}{(1+\text{t}^2)^2}$
$=\frac{4\text{a}^2\text{t}^2+\text{a}^2-2\text{a}^2\text{t}^2+\text{a}^2\text{t}^4}{(1+\text{t}^2)^2}$
$=\frac{2\text{a}^2\text{t}^2+\text{a}^2+\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}$
$=\frac{\text{a}^2(1+2\text{t}^2+\text{t}^2)}{(1+\text{t}^2)^2}$
$\text{x}^2+\text{y}^2=\text{a}^2$ is equation of a circle.

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Question 85 Marks

Find the equation of the circle which circumscribes the triangle formed by the lines
y = x + 2, 3y = 4x and 2y = 3x.

Answer

Solving equations y = x + 2 and 3y = 4x we get,
x = 6 and y = 8
Solving equations y = x + 2 and 2y = 3x we get,
x = 4 and y = 6
Solving equations 3y = 4x and 2y = 3x we get,
x = 0 and y = 0
So, the vertices of the triangle are (6, 8), ( 4, 6) and (0, 0).
Let x² + y² + 2gx + 2fy + c = 0 be the required cirde which circumscribes the triangle .
$\therefore$ x2 + y2 + 2gx + 2fy + c = 0 passes through ( 6, 8), (4, 6) and (0, 0).
12g + 16f + c = -100 ....... (1)
8g + 12f + C = -52 .......... (2)
C = 0 ....... (3)
Solving (i), (ii) and (iii) we get
f = 11 and g = -23
$\therefore$ The required equation of the circle is, x² + y² - 46x + 22y = 0.

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Question 95 Marks

Prove that the centres of the three circles x²+ y² - 4x - 6y - 12 = 0, x² + y² + 2x + 4y - 10 = 0 and x² + y² - 10x - 16y - 1 = 0 are collinear.

Answer

The given equation of circle are.
x² + y² - 4x - 6y - 12 = 0 .......... (1)
x² + y² + 2x + 4y - 10 = 0 ......... (2)
x² + y² - 10x - 16y - 1 = 0 ........ (3)
Let C₁ C₂ & C₃ are the centres of (1) (2) & (3)
$\therefore$ C₁= (-g, -f) = (2, 3)
C₂ = (-g, -f) = (-1, -2)
C₃ = (-g, -r) = (5, 8)
C₁ C₂ & C₃ will be collinear if ar $(\Delta\text{C}_1\ \text{C}_2\ \text{C}_3)=\bar0$
$(\Delta\text{C}_1\text{C}_2\text{C}_3)=\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}=\frac{1}{2}\begin{vmatrix}2&3&1\\-1&-2&1\\5&8&1\end{vmatrix}\\=\frac{1}{2}\begin{vmatrix}2&3&1\\-3&-5&0\\3&5&0\end{vmatrix}\begin{matrix}\text{R}_2\rightarrow\text{R}_2-\text{R}_3\\\text{R}_3\rightarrow\text{R}_3-\text{R}_1 \end{matrix}$
$=\frac{1}{2}(-15+15)=\frac{1}{2}\times0=0$
$\therefore$ C₁ C₂ & C₃ are collinear

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Question 105 Marks

Find the equation of the circle the end points of whose diameter are the centres of the circles x² + y² + 6x - 14y - 1 = 0 and x² + y² - 4x + 10y - 2 = 0.

Answer

Given:
x² + y² + 6x - 14y − 1 = 0 ........ (1)
And, x² + y² - 4x + 10y - 2 = 0 ...... (2)
Equations (1) and (2) can be rewritten as follows:
(x + 3)² + (y - 7)² = 59
And, (x - 2)² + (y + 5)² = 31
Thus, the centres of the circles are (-3, 7) and (2, -5).
Hence, the equation of the circle, the end points of whose diameter are the centres of the given circles, is (x + 3)(x - 2) + (y - 7)(y + 5) = 0, i.e.
x² + y² + x − 2y − 41 = 0.

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Question 115 Marks

Find the equation of the circle, the end points of whose diameter are (2, -3) and (-2, 4). Find its centre and radius.

Answer

(2, -3) and (-2, 4) are the ends points of the diameter of a circle. The equation of this
circle is (x - 2)(x + 2) + (y + 3)(y - 4) = 0.
⇒ x² - 4 + y²- 4y + 3y - 12 = 0
⇒ x² + y² - y - 16 = 0 ....... (1)
Equation (1) can be rewritten as
$\text{x}^2+\Big(\text{y}-\frac{1}{2}\Big)^2-\frac{1}{4}-16=0$
$\Rightarrow\text{x}^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\frac{65}{4}$
$\therefore$ Centre is $\Big(0,\ \frac{1}{2}\Big)$ and radius is $\frac{\sqrt{65}}{2}.$

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Question 125 Marks

A circle of radius 4 units touches the coordinate axes in the first quadrant. Find the equations of its images with respect to the line mirrors x = 0 and y = 0.

Answer

We are given that a circle has radius 4 and touches the coordinate axes in 1^st quadrant.
Thus the centre = ( 4, 4)
Now C₂ and C₃ are the images of C₁ with respect toy = 0 andx = 0
So, for C₂
Centre = (-4, 4) and radius = 4
Thus the equation of circle C2 is (x + 4)^₂ + (Y -4)² = 4²
⇒ x² + y² + 8x - 8y + 16 = 0
And for C₃
centre = ( 4, -4) and radius = 4
Thus, the equation of circle C₃ is (x - 4)² (y + 4)² = 4²
⇒ x² + y² - 8x + 8y + 16 = 0

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Question 135 Marks

If a circle passes through the point (0, 0),(a, 0),(0, b) then find the coordinates of its centre.

Answer

Consider the following figure.

In the above diagram, CA, CO and CB are equal
radii of the circle and hence we have,
CA = CO = CB = r.
Also the triangle $\Delta\text{OCA}$ is an isosceles triangle, and
CM is the perpendicular bisector to the base OA.
Hence $\text{CM}=\frac{\text{a}}{\text{2}}$
Similarly, CN is the perpendirular bisector to the base OB
Thus, $\text{ON}=2$
Thus, from the diagram it is dear that
$\text{CM}=\text{x}=\frac{\text{a}}{2}$
and
$\text{ON}=\text{y} =\frac{\text{b}}{2}$
Hence the centre of the circle is c $\Big(\frac{\text{a}}{2},\ \frac{\text{b}}{2}\Big)$

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Question 145 Marks

One diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If the coordinates of Aand B are (-3, 4) and (5, 4) respectively, find the equation of the circle.

Answer

The centre O Ii es on the I ine x - 4y = -7 and the perpendicular bi sector MO of AB.
The coordinates of Mare (1, 4).
Thus, the equation of MO is x = 1
Point of intersection of x - 4y = -7 and x = 1 is
0 = (1, 2)
Also the radius of ci rel e is
$\text{AO}=\sqrt{(1+3)^2+(2-4)^2}$
$=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$
Thus the equation of circle is
$(\text{x}-1)^2+(\text{y}-2)^2=20$
$\Rightarrow\text{x}^2+\text{y}^2-2\text{x}-4\text{y}-15=0$

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Question 155 Marks

Find the equation of the circle concentric with x² + y² - 4x - 6y - 3 = 0 and which touches the y-axis.

Answer

The given equabon of cir de is
x² + y² - 4x - 6y - 3 = 0 ......... (1)
$\therefore$ centre = (-9, -f) = (2, 3)
The required circle is oonc:entric with (1) so, they have same centre = (2, 3)
Also, the required cirde touchesy - axis at A
$\therefore$ CA = redius = 2
$\therefore$ equation of cirde is
(x - 2)² + (y - 3)² = 4
⇒ x² + y²-4x - 6y + 9 = 0

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Question 165 Marks

Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the x-axis and y-axis respectively.

Answer

We have a circle thet passes through origin O (0, 0) and cut of on mtersept of length 4 units on x-axis Be 6 units on y-axis.
That is, OA = 4
OB = 6
C - be the centre of the cirde and CM & CN are perpendicular line drawn on OA & 08 respectively.
Coordinates of A = ( 4, 0) & B = (0, 6)
$\therefore$ Coordinates of M = (2, 0) & N = (0, 3)
Thus coordinates of C = (2, 3)
Now in $\Delta\text{OCM}$
OC² = OM² + CM²
= 2² + 3² $[\because$ CM = ON =3 $]$
= 4 + 9
$\therefore\text{OC}=\sqrt{3}$
Thus, the required cirde is
(x - 2)²+ (y - 3)² = 13
x² + y² - 4X - 6y = 0

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Question 175 Marks

If the lines 3x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are tangents to a circle, then find the radius of the circle.

Answer

Slope of 3x - 4y + 4 = 0 is $\frac{4}{3}$
Slope of 6x - 8y -7 = 0 is $\frac{8}{6}=\frac{4}{3}$
Slope of 3x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are same.
Hence two lines are parallel and are shown in figure.

Rewriting 6x - 8y - 7 = 0, we get,
$\Bigg|\frac{4+\frac{7}{2}}{\sqrt{9+16}}\Bigg|$
$\Big|\frac{15}{10}\Big|$
$=\frac{3}{4}$ Units

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Question 185 Marks

If the line 2x - y + 1 = 0 touches the circle at the point (2, 5) and the centre of the circle lies on the line x + y - 9 = 0. Find the equation of the circle.

Answer

The line 2x - y + 1 = 0 touches the circle at A (2, 5). The centre of circle lies
on the line m : x + y = 9.
Now AO is perpendicular to 2x - y + 1 = 0
$\therefore$ equation of AO is
x + 2y = d ........ (3)
But AO passes through A (2, 5)
$\therefore$ d = 12
$\therefore$ equation of AO is
x + 2y = 12 ........ (4)
The point of intersection of x + y = 9 and x + 2y = 12 is (6, 3) which is the centre of the circle.
Radius $=\text{AO}=\sqrt{(6-2)^2(3-5)^2}=\sqrt{16+4}=\sqrt{20}$
Hence, equation of circle is
(x - 6)²+ (y - 3)² = 20

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Question 195 Marks

Find the equation of the circle whose diameter is the line segment joining (-4, 3) and (12, -1). Find also the intercept made by it on y-axis.

Answer

It is given that the end points of the diameter of the circle are (-4, 3) and (12, -1).
$\therefore$ Required equation of circle:
(x + 4)(x - 12) + (y - 3)(y + 1)
or x² + y²- 8x - 2y - 51 = 0 ...... (1)
Putting x = 0 in (1):
y² - 2y - 51 = 0
⇒ y² - 2y - 51 = 0
$\Rightarrow\text{y}=1\pm2\sqrt{3}$
Hence, the intercepts made by it on the y-axis is $1+2\sqrt{3}-1+2\sqrt{13}=4\sqrt{13}$

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Question 205 Marks

Find the equation of the circle which circumscribes the triangle formed by the lines
y = x + 2, 3y = 4x and 2y = 3x.

Answer

Solving equations y = x + 2 and 3y = 4x we get,
x = 6 and y = 8
Solving equations y = x + 2 and 2y = 3x we get,
x = 4 and y = 6
Solving equations 3y = 4x and 2y = 3x we get,
x = 0 and y = 0
So, the vertices of the triangle are (6, 8), ( 4, 6) and (0, 0).
Let x² + y² + 2gx + 2fy + c = 0 be the required cirde which circumscribes the triangle .
$\therefore$ x2 + y2 + 2gx + 2fy + c = 0 passes through ( 6, 8), (4, 6) and (0, 0).
12g + 16f + c = -100 ....... (1)
8g + 12f + C = -52 .......... (2)
C = 0 ....... (3)
Solving (i), (ii) and (iii) we get
f = 11 and g = -23
$\therefore$ The required equation of the circle is, x² + y² - 46x + 22y = 0.

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Question 215 Marks

Prove that the centres of the three circles x²+ y² - 4x - 6y - 12 = 0, x² + y² + 2x + 4y - 10 = 0 and x² + y² - 10x - 16y - 1 = 0 are collinear.

Answer

The given equation of circle are.
x² + y² - 4x - 6y - 12 = 0 .......... (1)
x² + y² + 2x + 4y - 10 = 0 ......... (2)
x² + y² - 10x - 16y - 1 = 0 ........ (3)
Let C₁ C₂ & C₃ are the centres of (1) (2) & (3)
$\therefore$ C₁= (-g, -f) = (2, 3)
C₂ = (-g, -f) = (-1, -2)
C₃ = (-g, -r) = (5, 8)
C₁ C₂ & C₃ will be collinear if ar $(\Delta\text{C}_1\ \text{C}_2\ \text{C}_3)=\bar0$
$(\Delta\text{C}_1\text{C}_2\text{C}_3)=\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}=\frac{1}{2}\begin{vmatrix}2&3&1\\-1&-2&1\\5&8&1\end{vmatrix}\\=\frac{1}{2}\begin{vmatrix}2&3&1\\-3&-5&0\\3&5&0\end{vmatrix}\begin{matrix}\text{R}_2\rightarrow\text{R}_2-\text{R}_3\\\text{R}_3\rightarrow\text{R}_3-\text{R}_1 \end{matrix}$
$=\frac{1}{2}(-15+15)=\frac{1}{2}\times0=0$
$\therefore$ C₁ C₂ & C₃ are collinear

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Question 225 Marks

Find the equation of the circle the end points of whose diameter are the centres of the circles x² + y² + 6x - 14y - 1 = 0 and x² + y² - 4x + 10y - 2 = 0.

Answer

Given:
x² + y² + 6x - 14y − 1 = 0 ........ (1)
And, x² + y² - 4x + 10y - 2 = 0 ...... (2)
Equations (1) and (2) can be rewritten as follows:
(x + 3)² + (y - 7)² = 59
And, (x - 2)² + (y + 5)² = 31
Thus, the centres of the circles are (-3, 7) and (2, -5).
Hence, the equation of the circle, the end points of whose diameter are the centres of the given circles, is (x + 3)(x - 2) + (y - 7)(y + 5) = 0, i.e.
x² + y² + x − 2y − 41 = 0.

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Question 235 Marks

The sides of a square are x = 6, x = 9, y = 3 and y = 6. Find the equation of a circle drawn on the diagonal of the square as its diameter.

Answer

Let the sides AB, BC, CD and DA of the square ABCD be represented by the equations
y = 3, x = 6, y = 6 and x = 9 respectively.
Then, coordinates are
A(6, 3), B(9, 3), C(9, 6) and D(6, 6).
The equation of the circle with diagonal AC
(x - 6)(x - 9) + (4 - 3)(4 - 6) = 0
⇒ x² - 6x - 9x + 54 + y² - 3y - 6y + 18 = 0
⇒ x² + y² - 15x - 9y + 72 = 0
The equation of the circle with diagonal BD as diameter is
(x - 9)(x - 6) + (y - 3) (y - 6) =
⇒ x² - 9x - 6x + 54 + y² - 3y - 6y + 18 = O
⇒ x² + y² - 15x - 9y + 72 = O
x² + y² - 15x - 9y + 72 = O

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Question 245 Marks

Find the equation of the circle, the end points of whose diameter are (2, -3) and (-2, 4). Find its centre and radius.

Answer

(2, -3) and (-2, 4) are the ends points of the diameter of a circle. The equation of this
circle is (x - 2)(x + 2) + (y + 3)(y - 4) = 0.
⇒ x² - 4 + y²- 4y + 3y - 12 = 0
⇒ x² + y² - y - 16 = 0 ....... (1)
Equation (1) can be rewritten as
$\text{x}^2+\Big(\text{y}-\frac{1}{2}\Big)^2-\frac{1}{4}-16=0$
$\Rightarrow\text{x}^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\frac{65}{4}$
$\therefore$ Centre is $\Big(0,\ \frac{1}{2}\Big)$ and radius is $\frac{\sqrt{65}}{2}.$

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Question 255 Marks

Find the equation of the circle which passes through the points (2, 3) and (4,5) and the centre lies on the straight line y - 4x + 3 = 0

Answer

Find the equation of the circle which passes through the point (2, 3) and ( 4, 5)
and the centre lies on the straight line y - 4x + 3 = 0.
Let the equation of required circle be x² + y² + 2gx+ 2fy + c = 0 which passes
through the point (2, 3) and ( 4, 5) .
$\therefore$ 13 + 4g + 6f + c = 0 ......... (1)
41 + 8g +10f + c = 0 ........... (2)
Centre (-g, -f) lies on y - 4x + 3 = 0
-f + 4g = -3 .......... (3)
Subtracting (1) from (2), we get
28 + 4g + 4f = 0 ........ (4)
Solving (3) and (4) we get,
f = -5 and g = -2
Substituting values off and g in (2) we get,
41 - 16 - 50 + c = 0
c = 25
$\therefore$ The required equation of the circle is, x² + y² - 4x - 10y + 25 = 0.

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Question 265 Marks

Find the equations of the circles touching y-axis at (0, 3) and making an intercept of 8 units on the x-axis.

Answer

Case I: The centre lies in first quadrant.

Let the required equation be (x - h)² + (y - k)² = a²

Here, AB = 8 units and L (0,

In $\triangle\text{CAM}$

⇒ CA² = CM² + AM²

⇒ CA² = 3² + 4²

⇒ CA = 5

⇒ CL = CA = 5

$\therefore$ Coordinates of the centre = (5, 3)

And, radius of the circle = 5

(x - 5)² + (y - 3)² = 25, i.e. x² + y² - 10x - 6y= -9

Case II: The centre lies in the second quadrant.

Coordinates of the centre = (-5, 3)

And, radius of the circle = 5

(x - 5)² + (y - 3)² = 25, i.e. x² + y² - 10x - 6y= -9

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Question 275 Marks

A circle of radius 4 units touches the coordinate axes in the first quadrant. Find the equations of its images with respect to the line mirrors x = 0 and y = 0.

Answer

We are given that a circle has radius 4 and touches the coordinate axes in 1^st quadrant.
Thus the centre = ( 4, 4)
Now C₂ and C₃ are the images of C₁ with respect toy = 0 andx = 0
So, for C₂
Centre = (-4, 4) and radius = 4
Thus the equation of circle C2 is (x + 4)^₂ + (Y -4)² = 4²
⇒ x² + y² + 8x - 8y + 16 = 0
And for C₃
centre = ( 4, -4) and radius = 4
Thus, the equation of circle C₃ is (x - 4)² (y + 4)² = 4²
⇒ x² + y² - 8x + 8y + 16 = 0

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Question 285 Marks

One diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If the coordinates of Aand B are (-3, 4) and (5, 4) respectively, find the equation of the circle.

Answer

The centre O Ii es on the I ine x - 4y = -7 and the perpendicular bi sector MO of AB.
The coordinates of Mare (1, 4).
Thus, the equation of MO is x = 1
Point of intersection of x - 4y = -7 and x = 1 is
0 = (1, 2)
Also the radius of ci rel e is
$\text{AO}=\sqrt{(1+3)^2+(2-4)^2}$
$=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$
Thus the equation of circle is
$(\text{x}-1)^2+(\text{y}-2)^2=20$
$\Rightarrow\text{x}^2+\text{y}^2-2\text{x}-4\text{y}-15=0$

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Question 295 Marks

Find the equations of the circles touching y-axis at (0, 3) and making an intercept of 8 units on the x-axis.

Answer

Case I: The centre lies in first quadrant.

Let the required equation be (x - h)² + (y - k)² = a²

Here, AB = 8 units and L (0,

In $\triangle\text{CAM}$

⇒ CA² = CM² + AM²

⇒ CA² = 3² + 4²

⇒ CA = 5

⇒ CL = CA = 5

$\therefore$ Coordinates of the centre = (5, 3)

And, radius of the circle = 5

(x - 5)² + (y - 3)² = 25, i.e. x² + y² - 10x - 6y= -9

Case II: The centre lies in the second quadrant.

Coordinates of the centre = (-5, 3)

And, radius of the circle = 5

(x - 5)² + (y - 3)² = 25, i.e. x² + y² - 10x - 6y= -9

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Question 305 Marks

Find the equation of the circle concentric with x² + y² - 4x - 6y - 3 = 0 and which touches the y-axis.

Answer

The given equabon of cir de is
x² + y² - 4x - 6y - 3 = 0 ......... (1)
$\therefore$ centre = (-9, -f) = (2, 3)
The required circle is oonc:entric with (1) so, they have same centre = (2, 3)
Also, the required cirde touchesy - axis at A
$\therefore$ CA = redius = 2
$\therefore$ equation of cirde is
(x - 2)² + (y - 3)² = 4
⇒ x² + y²-4x - 6y + 9 = 0

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Question 315 Marks

Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the x-axis and y-axis respectively.

Answer

We have a circle thet passes through origin O (0, 0) and cut of on mtersept of length 4 units on x-axis Be 6 units on y-axis.
That is, OA = 4
OB = 6
C - be the centre of the cirde and CM & CN are perpendicular line drawn on OA & 08 respectively.
Coordinates of A = ( 4, 0) & B = (0, 6)
$\therefore$ Coordinates of M = (2, 0) & N = (0, 3)
Thus coordinates of C = (2, 3)
Now in $\Delta\text{OCM}$
OC² = OM² + CM²
= 2² + 3² $[\because$ CM = ON =3 $]$
= 4 + 9
$\therefore\text{OC}=\sqrt{3}$
Thus, the required cirde is
(x - 2)²+ (y - 3)² = 13
x² + y² - 4X - 6y = 0

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Question 325 Marks

If the lines 3x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are tangents to a circle, then find the radius of the circle.

Answer

Slope of 3x - 4y + 4 = 0 is $\frac{4}{3}$
Slope of 6x - 8y -7 = 0 is $\frac{8}{6}=\frac{4}{3}$
Slope of 3x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are same.
Hence two lines are parallel and are shown in figure.

Rewriting 6x - 8y - 7 = 0, we get,
$\Bigg|\frac{4+\frac{7}{2}}{\sqrt{9+16}}\Bigg|$
$\Big|\frac{15}{10}\Big|$
$=\frac{3}{4}$ Units

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Question 335 Marks

Find the equation of the circle whose diameter is the line segment joining (-4, 3) and (12, -1). Find also the intercept made by it on y-axis.

Answer

It is given that the end points of the diameter of the circle are (-4, 3) and (12, -1).
$\therefore$ Required equation of circle:
(x + 4)(x - 12) + (y - 3)(y + 1)
or x² + y²- 8x - 2y - 51 = 0 ...... (1)
Putting x = 0 in (1):
y² - 2y - 51 = 0
⇒ y² - 2y - 51 = 0
$\Rightarrow\text{y}=1\pm2\sqrt{3}$
Hence, the intercepts made by it on the y-axis is $1+2\sqrt{3}-1+2\sqrt{13}=4\sqrt{13}$

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Question 345 Marks

Find the equation of the circle which circumscribes the triangle formed by the lines
x + y = 2, 3x - 4y = 6 and x - y = 0.

Answer

The given equation of lines
x + y = 2 ......... (1)
3x - 4y = 6 ....... (2)
x - y = 0 ........... (3)
Let A, B & C are the point of intersection of lines (1) & (2), (2) & (3) and (3) & (1) respectively
$\therefore$ A = (2, 0), B = (-6, -6) & C = (1, 1)
Let x² + y²+ 2gx + 2fy + c = 0 ........ (A)
be the circle that circum scnbinq $\Delta\text{ABC}$
$\therefore$ 4 + 4g + c = 0 ........ (4)
36 + 36 + 12g + 12f + c = 0 ........ (5)
1 + 1 + 2g + 2f + c = 0 ....... (4)
Solving (4), (5) & (6) we get,
g = 2, f = 3 & c = -12
from (A),
The required cirde is
x² + y² - 4x + 6y - 12 = 0

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Question 355 Marks

Show that the point $(\text{x},\ \text{y})$ given by $\text{x}=\frac{2\text{at}}{1+\text{t}^2}$ and $\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}^2}\Big)$2 lies on a circle for all real values of t such that $-1\leq\text{t}\leq1,$ where a is any given real number.

Answer

$\text{x}=\frac{2\text{at}}{1+\text{t}^2},\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}}\Big)$
$\text{x}^2+\text{y}^2=\frac{4\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}+\frac{\text{a}^2(1-\text{t}^2)^2}{(1+\text{t}^2)^2}$
$=\frac{4\text{a}^2\text{t}^2+\text{a}^2(1-2\text{t}^2+\text{a}^2)\text{t}^4}{(1+\text{t}^2)^2}$
$=\frac{4\text{a}^2\text{t}^2+\text{a}^2-2\text{a}^2\text{t}^2+\text{a}^2\text{t}^4}{(1+\text{t}^2)^2}$
$=\frac{2\text{a}^2\text{t}^2+\text{a}^2+\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}$
$=\frac{\text{a}^2(1+2\text{t}^2+\text{t}^2)}{(1+\text{t}^2)^2}$
$\text{x}^2+\text{y}^2=\text{a}^2$ is equation of a circle.

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Question 365 Marks

Find the equations of the circles passing through two points on y-axis at distances 3 from the origin and having radius 5.

Answer

Let the required equation of the circle be (x - h)² + (y - k)² = a²
The circle passes through the points (0, 3) and (0, -3).
$\therefore$ (0 - h)² + (3 - k)² = a2 ........ (1)
And, (0 - h)² + (-3 - k)² =a² ....... (2)
Solving (1) and (2), we get:
k = 0
Given:
Radius = 5
$\therefore$ a² = 25
So, from equation (2), we have:
h² + 9 = 25 ⇒ h = ±4
Hence, the required equation is (x ± 4)² + y² = 25, which can be rewritten as
x² ± 8x + y² − 9 = 0

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Question 375 Marks

If the line 2x - y + 1 = 0 touches the circle at the point (2, 5) and the centre of the circle lies on the line x + y - 9 = 0. Find the equation of the circle.

Answer

The line 2x - y + 1 = 0 touches the circle at A (2, 5). The centre of circle lies
on the line m : x + y = 9.
Now AO is perpendicular to 2x - y + 1 = 0
$\therefore$ equation of AO is
x + 2y = d ........ (3)
But AO passes through A (2, 5)
$\therefore$ d = 12
$\therefore$ equation of AO is
x + 2y = 12 ........ (4)
The point of intersection of x + y = 9 and x + 2y = 12 is (6, 3) which is the centre of the circle.
Radius $=\text{AO}=\sqrt{(6-2)^2(3-5)^2}=\sqrt{16+4}=\sqrt{20}$
Hence, equation of circle is
(x - 6)²+ (y - 3)² = 20

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Question 385 Marks

Find the equation to the circle which passes through the points (1, 1) (2, 2) and whose radius is 1. Show that there are two such circles.

Answer

Let x² + y² + 2gx + 2fy + c = 0 ........(1)
be the required cirde.
Now (1) passes through $\text{P}=(1,\ 1)\&\ \theta(2,\ 2)$
$\therefore$ 1 + 1 + 2g + 2f + C = 0 ........... (2)
4 + 4 + 4g + 4f + c = 0 ............... (3)
Also racius = 1
$\Rightarrow\sqrt{\text{g}^2+\text{f}^2-\text{c}}=1$
$\Rightarrow\text{g}^2+\text{f}^2-\text{c}1\ .......(4)$
from (2) & (4)
$\text{g}+\text{f}+\frac{\text{c}}{2}=-1\ \&$
$\text{g}+\text{f}+\frac{\text{c}}{4}=-2$
on subtraction
and $\text{g}+\text{f}=-3\ ........(4)$
From (4) $\text{g}^2+\text{f}^2=5$ $\big\{\therefore(\text{g}+\text{f})^2=\text{g}^2+\text{f}^2+2\text{g}\text{f}\big\}$
$\therefore2\text{gf}=4$ $\Rightarrow9=5+2\text{gf}$
$\text{gf}=2$
so, $(\text{g}-\text{f})^2=\text{g}+\text{f})^2-4\text{gf}=9-8=1$
$\therefore\text{g}-\text{f}=\pm1\ .........\ (4)$
Solving (5) & (6) we get
$\text{g}=-1$ or $-2\ \&$
$\text{f}=-2$ or $-1$
Thus, required circle
$\text{x}^2+\text{y}^2-2\text{x}-4\text{y}+4=0$
$\text{x}^2+\text{y}^2-4\text{x}-2\text{y}+4=0$

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Question 395 Marks

The sides of a square are x = 6, x = 9, y = 3 and y = 6. Find the equation of a circle drawn on the diagonal of the square as its diameter.

Answer

Let the sides AB, BC, CD and DA of the square ABCD be represented by the equations
y = 3, x = 6, y = 6 and x = 9 respectively.
Then, coordinates are
A(6, 3), B(9, 3), C(9, 6) and D(6, 6).
The equation of the circle with diagonal AC
(x - 6)(x - 9) + (4 - 3)(4 - 6) = 0
⇒ x² - 6x - 9x + 54 + y² - 3y - 6y + 18 = 0
⇒ x² + y² - 15x - 9y + 72 = 0
The equation of the circle with diagonal BD as diameter is
(x - 9)(x - 6) + (y - 3) (y - 6) =
⇒ x² - 9x - 6x + 54 + y² - 3y - 6y + 18 = O
⇒ x² + y² - 15x - 9y + 72 = O
x² + y² - 15x - 9y + 72 = O

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Question 405 Marks

If a circle passes through the point (0, 0),(a, 0),(0, b) then find the coordinates of its centre.

Answer

Consider the following figure.

In the above diagram, CA, CO and CB are equal
radii of the circle and hence we have,
CA = CO = CB = r.
Also the triangle $\Delta\text{OCA}$ is an isosceles triangle, and
CM is the perpendicular bisector to the base OA.
Hence $\text{CM}=\frac{\text{a}}{\text{2}}$
Similarly, CN is the perpendirular bisector to the base OB
Thus, $\text{ON}=2$
Thus, from the diagram it is dear that
$\text{CM}=\text{x}=\frac{\text{a}}{2}$
and
$\text{ON}=\text{y} =\frac{\text{b}}{2}$
Hence the centre of the circle is c $\Big(\frac{\text{a}}{2},\ \frac{\text{b}}{2}\Big)$

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Question 415 Marks

A circle whose centre is the point of intersection of the lines 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0 passes through the origin. Find its equation.

Answer

we heve,
$2\text{x}-3\text{y}=-4\ .......(1)$
$3\text{x}+4\text{y}=5\ .........(2)$
The point of intersection of (1) & (2) is
$\text{p}\Big(\frac{-1}{17},\ \frac{66}{51}\Big)$ or $\text{p}\Big(\frac{-1}{17},\ \frac{22}{17}\Big)$
According to the equation centre $=\Big(\frac{-1}{17},\ \frac{22}{17}\Big)$
Also, the cirde passes through 0 (0, 0)
$\therefore\ \text{r}=\text{OC}=\sqrt{\Big(0+\frac{1}{17}\Big)^2+\Big(0-\frac{22}{17}\Big)^2}$
$=\sqrt{\frac{1}{289}+\frac{484}{289}}=\frac{\sqrt{485}}{17}$
Thus, the required equation of circle is
$\Big(\text{x}+\frac{1}{17}\Big)^2+\Big(\text{y}-\frac{22}{17}\Big)^2=\frac{485}{289}$

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Question 425 Marks

Find the equation of the circle concentric with the circle x² + y² - 6x + 12y + 15 = 0 and double of its area.

Answer

The given equation of cirde is
x²+ y² - 6x + 12y + 15 = o ......... (1)
$\therefore$ centre= (-g, -1) = (3,-6)
Radius $\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{9+36-15}=\sqrt{30}$
Now,
The required equation of cirde in coocenmc with(1)
Which means both have same centre (3, -6)
Also,
Area of required cirde = 2 × Area of (1)
$\pi\text{r}^2=2\times\pi(\sqrt{30})^2$
$\Rightarrow\text{R}^2=60$
$\Rightarrow\text{R}=2\sqrt{15}$
Thus,
The required circle is
(x-3)² + (y +6)²=60
x² + y² - 6x +12y - 15 = O

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Question 435 Marks

Find the equation of the circle which circumscribes the triangle formed by the lines
x + y = 2, 3x - 4y = 6 and x - y = 0.

Answer

The given equation of lines
x + y = 2 ......... (1)
3x - 4y = 6 ....... (2)
x - y = 0 ........... (3)
Let A, B & C are the point of intersection of lines (1) & (2), (2) & (3) and (3) & (1) respectively
$\therefore$ A = (2, 0), B = (-6, -6) & C = (1, 1)
Let x² + y²+ 2gx + 2fy + c = 0 ........ (A)
be the circle that circum scnbinq $\Delta\text{ABC}$
$\therefore$ 4 + 4g + c = 0 ........ (4)
36 + 36 + 12g + 12f + c = 0 ........ (5)
1 + 1 + 2g + 2f + c = 0 ....... (4)
Solving (4), (5) & (6) we get,
g = 2, f = 3 & c = -12
from (A),
The required cirde is
x² + y² - 4x + 6y - 12 = 0

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Question 445 Marks

A circle whose centre is the point of intersection of the lines 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0 passes through the origin. Find its equation.

Answer

we heve,
$2\text{x}-3\text{y}=-4\ .......(1)$
$3\text{x}+4\text{y}=5\ .........(2)$
The point of intersection of (1) & (2) is
$\text{p}\Big(\frac{-1}{17},\ \frac{66}{51}\Big)$ or $\text{p}\Big(\frac{-1}{17},\ \frac{22}{17}\Big)$
According to the equation centre $=\Big(\frac{-1}{17},\ \frac{22}{17}\Big)$
Also, the cirde passes through 0 (0, 0)
$\therefore\ \text{r}=\text{OC}=\sqrt{\Big(0+\frac{1}{17}\Big)^2+\Big(0-\frac{22}{17}\Big)^2}$
$=\sqrt{\frac{1}{289}+\frac{484}{289}}=\frac{\sqrt{485}}{17}$
Thus, the required equation of circle is
$\Big(\text{x}+\frac{1}{17}\Big)^2+\Big(\text{y}-\frac{22}{17}\Big)^2=\frac{485}{289}$

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Question 455 Marks

Find the equation to the circle which passes through the points (1, 1) (2, 2) and whose radius is 1. Show that there are two such circles.

Answer

Let x² + y² + 2gx + 2fy + c = 0 ........(1)
be the required cirde.
Now (1) passes through $\text{P}=(1,\ 1)\&\ \theta(2,\ 2)$
$\therefore$ 1 + 1 + 2g + 2f + C = 0 ........... (2)
4 + 4 + 4g + 4f + c = 0 ............... (3)
Also racius = 1
$\Rightarrow\sqrt{\text{g}^2+\text{f}^2-\text{c}}=1$
$\Rightarrow\text{g}^2+\text{f}^2-\text{c}1\ .......(4)$
from (2) & (4)
$\text{g}+\text{f}+\frac{\text{c}}{2}=-1\ \&$
$\text{g}+\text{f}+\frac{\text{c}}{4}=-2$
on subtraction
and $\text{g}+\text{f}=-3\ ........(4)$
From (4) $\text{g}^2+\text{f}^2=5$ $\big\{\therefore(\text{g}+\text{f})^2=\text{g}^2+\text{f}^2+2\text{g}\text{f}\big\}$
$\therefore2\text{gf}=4$ $\Rightarrow9=5+2\text{gf}$
$\text{gf}=2$
so, $(\text{g}-\text{f})^2=\text{g}+\text{f})^2-4\text{gf}=9-8=1$
$\therefore\text{g}-\text{f}=\pm1\ .........\ (4)$
Solving (5) & (6) we get
$\text{g}=-1$ or $-2\ \&$
$\text{f}=-2$ or $-1$
Thus, required circle
$\text{x}^2+\text{y}^2-2\text{x}-4\text{y}+4=0$
$\text{x}^2+\text{y}^2-4\text{x}-2\text{y}+4=0$

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Question 465 Marks

Find the equation of the circle concentric with the circle x² + y² - 6x + 12y + 15 = 0 and double of its area.

Answer

The given equation of cirde is
x²+ y² - 6x + 12y + 15 = o ......... (1)
$\therefore$ centre= (-g, -1) = (3,-6)
Radius $\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{9+36-15}=\sqrt{30}$
Now,
The required equation of cirde in coocenmc with(1)
Which means both have same centre (3, -6)
Also,
Area of required cirde = 2 × Area of (1)
$\pi\text{r}^2=2\times\pi(\sqrt{30})^2$
$\Rightarrow\text{R}^2=60$
$\Rightarrow\text{R}=2\sqrt{15}$
Thus,
The required circle is
(x-3)² + (y +6)²=60
x² + y² - 6x +12y - 15 = O

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