1 Marks Question

Question 11 Mark

Write the area of the circle passing through (-2, 6) and having its centre at (1, 2).

Answer

Radius of circle = Distance between centre (1, 2) and (-2, 6)
$=\sqrt{(1+2)^2+(2-6)^2}$
$=\sqrt{9+16}$
$=5$
Area of the circle = $25\pi$

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Question 21 Mark

Write the coordinates of the centre of the circle inscribed in the square formed by the lines x = 2, x = 6, y = 5 and y = 9.

Answer

Cir de is inscribed in the square formed by lines x = 2, x = 6 and y = 5, y = 9
It is a square of side 4
Thus, abscissae of centre $=\frac{2+6}{2}=4$
Ordinate of centre $=\frac{5+9}{2}=7$
Centre of the circle = (4, 7)

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Question 31 Mark

Find the equation of the circle with:
Centre (a, b) and radius $\sqrt{\text{a}^2+\text{b}^2}$

Answer

The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b)are centre and r is radius
From (A)
$(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\big(\sqrt{\text{a}^2+\text{b}}\big)^2$
$\Rightarrow\text{x}^2-2\text{ax}+\text{a}^2+\text{y}^2-2\text{bx}+\text{b}^2=\text{a}^2+\text{b}^2$
$\Rightarrow\text{x}^2+\text{y}^2-2\text{ax}-2\text{by}=0$

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Question 41 Mark

If the radius of the circle x² + y² + ax + (1 - a) y + 5 = 0 does not exceed 5, write the number of integral values a.

Answer

Equation of given circle is x² + y² + ax + (1 - a)y + 5 = 0
$\text{g}=\frac{\text{a}}{\text{2}},\ \text{f}=\frac{(1-\text{a})}{2}$
Radius $=\text{r}=\sqrt{\text{g}^2+\text{f}^2-\text{c}}$
$=\sqrt{\frac{\text{a}^2}{4}}+\frac{(1-\text{a}^2)}{4}-5<5$ .........[Radius can be at most]
$\frac{\text{a}^2}{4}+\frac{(1-\text{a})^2}{4}-5<25$
$\text{a}^2+(1-\text{a}^2)<120$
Sum of two squares should be I ess than 120.
a can be at most 8 and atleast -7.
So a can take 16 integral values which are from -7 to 8

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Question 51 Mark

If the line y = mx does not intersect the circle (x + 10)² + (y + 10)² = 180, then write the set of values taken by m.

Answer

y = mx does not intersect the circle (x + 10)² + (y + 10)² = 180
Now, (x + 10)²+ (mx + 10)² = 180
x² +100 + 20x + m²x² + 100 + 20mx = 180
x² + m²x² + 20mx + 20x + 20 = 0
x² (1 + m²) + 20x (m + 1) + 20 = 0
Since y = mx does not intersect so
b² - 4ac < 0
[20(m + 1)² ] -4(1 + m²)(20)< 0
400 (m + 1)² - 80 (1 + m²) < 0
5m² + 5 + 10m - 1 - m² < 0
4m² + 10m + 4 < 0
4m² + 8m + 2m + 4 < 0
4m(m + 2) + 2 (m + 2) < 0
(m + 2)(4m + 2) < 0
⇒ m + 2 < 0 and 4m + 2 > 0
⇒ m < -2 and m > -2
Its not possible
$\Rightarrow\text{m}+2>0$ and $4\text{m}+2<0$
$\Rightarrow\text{m}>-2$ and $\text{m}<-\frac{1}{2}$
$\Rightarrow-2<\text{m}<-\frac{1}{2}$
$\Rightarrow\text{m}\in\Big(-2,\ \frac{1}{2}\Big)$

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Question 61 Mark

Write the length of the intercept made by the circle x² + y² + 2x - 4y - 5 = 0 on y-axis.

Answer

$\text{x}^2+\text{y}^2+2\text{x}-4\text{y}-5=0$
Put x = 0
$\text{y}^2-4\text{y}-5=0$
$\text{y}^2-5\text{y}+\text{y}-5=0$
$\text{y}(\text{y}-5)+(\text{y}-5)=0$
$(\text{y}-5)(\text{y}+1)=0$
$\text{y}=5,-1$
Thus, circle cuts y-axis at (0, 5) and (0, -1) so
length of intercept on y-axisby circle = 5 + 1 = 6 units.

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Question 71 Mark

Find the equation of the circle with:
Centre $(\text{a}\cos\alpha,\ \text{a}\sin\alpha)$ and radius a.

Answer

The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b)are centre and r is radius
From (A)
$(\text{x}-\text{a}\cos\alpha)^2+(\text{y}-\text{a}\sin\alpha)^2=\text{a}^2$
$\Rightarrow\text{x}^2-2\text{a}\cos\alpha\text{x}-2\text{a}\sin\alpha\ \text{y}=0$

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Question 81 Mark

If the abscissae and ordinates of two points P and Q are roots of the equations x² + 2ax - b² = 0 and x² + 2px - q² = 0 respectively, then write the equation of the circle with PQ as diameter.

Answer

Let $\alpha,\ \beta$ abd $\gamma,\ \delta$ are the roots of the first and second given equation, so
$\alpha+\beta=-2\text{a}$ $\alpha\beta=-\text{b}^2$
$\gamma+\delta=-2\text{p}\ ,\ \gamma\delta=-\text{q}^2$
Coordinate of p and q are $(\alpha,\ \gamma)$ and $(\beta,\ \delta)$ respectively the equation of circle on PQ as diameter is
$(\text{x}-\alpha)(\text{x}-\beta)+(\text{y}-\gamma)(\text{y}-\delta)=0$
$\text{x}^2+\text{y}^2-(\alpha-\beta)\times-(\gamma+\delta)\text{y}+\alpha\beta+\gamma\beta=0$
$\text{x}^2+\text{y}^2+2\text{ax}+2\text{py}-\text{b}^2-\text{q}^2=0$

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Question 91 Mark

Find the equation of the circle with:
Centre (0, -1) and radius 1.

Answer

The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b) are centre and r is radius
From (A)
$(\text{x}-0)^2+(\text{y}+1)^2=1^2$
$\Rightarrow\text{x}^2+\text{y}^2+2\text{y}+1=1$
$\Rightarrow\text{x}^2+\text{y}^2+2\text{y}=0$

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Question 101 Mark

Write the equation of the unit circle concentric with x² + y² - 8x + 4y - 8 = 0.

Answer

x² + y² - Bx + 4y - 8 = 0
x² - Bx + (4)² - (4)² + y² + 4y + (2)² - (2)² - 8 = 0
(x - 4)² + (y + 2)² - 16 - 4 - 8 = 0
(x - 4) + (y + 2) - 28 = 0
(x - 4)² + (y + 2)² = 28 .......... (1)
Concentric unit circle to equation (1) is
(x - 4)² + (y + 2)² = 1
x² - Bx + y² + 4y + 19 = 0

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Question 111 Mark

Write the coordinates of the centre of the circle passing through (0, 0), (4, 0) and (0, -6).

Answer

Circle is passing through (0, 0), (4, 0) and (0, −6).
Let equation of circle be,
(x - h)² + (y - k)² = r² ...(1)
Circle (1) is passing through A (0 , 0)
(0 - h)² + (0 - k)² = r²
h² + k² = r² ...(2)
Circle (1) is passing through B (4 , 0)
(4 - h)2 + (0 - k)2 = r2
(4 - h)² + k² = r² ...(3)
Circle (1) is passing through B (0, -6)
(0 - h)² + (-6 - k)² = r²
h² + (6 + k)² = r² ...(4)
[(2) - (3)],
h² - (4 - h)² = 0
(h - 4 + h) (h + 4 - h) = 0
(2h - 4) (4) = 0
h = 2
[(2) - (4)],
k² - (6 + k)² = 0
(k - 4 - k) (k + 6 + k) = 0
(-6) (2k + 6 = 0)
k = - 3
put the value of h and k in equation (1)
h² + k² = r²
(2)² + (-3)² = r²
r² = 13
Centre of the circle = (2, - 3)

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Question 121 Mark

Find the equation of the circle with:
Centre (-2, 3) and radius 4.

Answer

The general equation of circle is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}\ .....(\text{A})$
where(a, b)are centre and r is radius
$\therefore(\text{x}+2)^2+(\text{y}-3)^2=4^2$
$\Rightarrow(\text{x}+2)^2+(\text{y}-3)^2=16$

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Question 131 Mark

Find the centre and radius of the following circles:
$(\text{x}-1)^2+\text{y}^2=4$

Answer

lhe general equation of orde is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2-\text{r}^2$
or $\text{x}^2+\text{y}^2-2\text{ax}-2\text{by}+\text{a}^2+\text{b}^2=\text{r}^2\ .....(\text{A})$
Where (a, b),s the centre and r be the radius of the circle.
$(\text{x}-1)^2+\text{y}^2=4$
$(\text{x}-1)^2+\text{y}^2-4$
$\Rightarrow(\text{x}-1)^2+(\text{y}-0)^2-2^2$
comparing with (A) we get,
(1. 0) is the centre
2 ts the radius

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Question 141 Mark

Find the equation of the circle with:
Centre (a, a) and radius $\sqrt{2}$ a.

Answer

The general equation of cirde is $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ .....(\text{A})$
where(a, b)are centre and r is radius
From (A)
$(\text{x}-\text{a})^2+(\text{y}-\text{a})^2=(\sqrt{2}\text{a})^2$
$\Rightarrow\text{x}^2-2\text{ax}+\text{a}^2+\text{y}^2-2\text{ay}+\text{a}^2=2\text{a}^2$
$\Rightarrow\text{x}^2+\text{y}^2-2\text{ax}-2\text{ay}=0$

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Question 151 Mark

Write the equation of the circle passing through (3, 4) and touching y-axis at the origin.

Answer

Cirde is touching y - axis at the origin. Thus, centre of cirde is on x- axis. So

Let centre of d rcle is ( h, o)

Equation of circle is

(x - h)² + (y - o)² = r² ............ (1)

It is passing through (3, 4)

(3 - h)² + (4)² = r²

(3 - h)² + 16 = r² ............ (2)

It is also passing through (3, 4)

(0 - h)² + (0 - o)² = r²

h² = r²

Now, equation ( 2) be com es,

(3 - h)² + 16 = h²

9 - 6h + h² + 16

-6h + 25 = 0

$\text{h}=\frac{25}{6}$

Equation of circle is,

(x - h)² + y² = h²

x² - 2xh + y² = 0

$\text{x}^2-2\text{x}\Big(\frac{25}{6}\Big)+\text{y}^2=0$

6x² - 50x + 6y² = 0

3x² - 25x + 3y² = 0

3(x² + y²) - 25x = 0

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