3 Marks Question - MATHS STD 11 Science Questions - Vidyadip

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Question 13 Marks

Find the equation of the circle which touches the axes and whose centre lies on x - 2y = 3.

Answer

If the circle lies in the third quadrant, then its centre will be (-a, -a).
The centre lies on x - 2y = 3.
$\therefore$ -a + 2a = 3 ⇒ a = 3
$\therefore$ Required equation of the circle = (x + 3)² + (y + 3)² = 9
= x² + y² + 6x + 6y + 9 = 0
If the circle lies in the fourth quadrant, then its centre will be (a, -a),
$\therefore$ a + 2a = 3 ⇒ a = 1
$\therefore$ Required equation of the circle = (x - 1)² + (y + 1)² = 1
= x² +y² − 2x + 2y + 1 = 0

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Question 23 Marks

Find the equation of the circle passing through the points:
(5, 7), (8, 1) and (1, 3)

Answer

We know that the general equation of cirde is x² + y²+ 2gx + 2fy + c = 0 ......... (1)
we have,
P (5, 7), Q (8, 1) and R (1, 3)
Since P, Q and R lies on (1)
so,
25 + 49 + 10g + 14f + C = 0 .......... (2)
64 + 1 + 16g + 2f +c = 0 ............... (3)
1 + 9 + 2g + 6f + c = 0 .................. (4)
Solving (2), (3) & (4), we get,
$\text{g}=-\frac{29}{6},\ \text{f}=\frac{19}{6},\ \text{c}=\frac{56}{3}$
Thus, the equation of circle is on putting g, f & con (1)
$\text{x}^2+\text{y}^2-\frac{29}{3}\times-\frac{19}{3}\text{y}+\frac{56}{3}=0$
$\Rightarrow3(\text{x}^2+\text{y}^2)-29\text{x}-19\text{y}+56=0$

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Question 33 Marks

Find the equation of the circle passing through the points:
(0, 0), (-2, 1) and (-3, 2)

Answer

We know that the general equation of circle is x² + y² + 2gx + 2fy + c = 0 ........ (1)
We have,
P (0, 0), Q (-2, 1) and R (-3, 2)
P, Q & R lies on(1), so,
0 + 00 + 0g + c = 0 ............ (2)
4 + 1 + 4x - 2y + c = 0 ........ (3)
9 + 4 + 6x - 4y + c = 0 ........ (4)
Solving (2), (3) & (4), we get,
$\text{g}=-\frac{3}{2},\ \text{f}=-\frac{11}{2}=-\frac{11}{2},\ \text{c}=0$
from (1),
Thus, equation of eirel e is,
x² + y² + 3x - 11y = 0

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Question 43 Marks

Find the equation of a circle,
Passing through the origin, radius 17 and ordinate of the centre is -15.

Answer

The circle passes through origin (0, 0) and has radius = 17 units
Also, the ordinate of centre is -15 then assume abssisa is a.
$\therefore\text{OC}=17$
$\Rightarrow\sqrt{(\text{a}-0)^2+(0+15)^2}=17$ (by distance form ulla)
$\Rightarrow\sqrt{\text{a}^2+225}=17$
$\Rightarrow\text{a}^2+225=289$
$\Rightarrow\text{a}^2=64$
$\Rightarrow\text{a}-\pm8$
$\therefore$ Centre $=(\pm8,\ -15)$
Thus, the equation of circle will be,
$(\text{x}\pm8)^2+(\text{y}+15)^2=17^2$
$\Rightarrow\text{x}^2+\text{y}^2\mp16\text{x}+30\text{y}=0$

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Question 53 Marks

Find the equation of a circle,
Which touches x-axis at a distance 5 from the origin and radius 6 units.

Answer

The circle touches the x-axis at A = (5, 0) and has radius 6 unit
Thus,
centre = (5, b)
By distance form ulla OA = 6
$\Rightarrow\sqrt{(5-5)^2+(\text{b}-0)^2}=6$
$\Rightarrow\text{b}=6$
$\Rightarrow$ Centre $=(5,\ 6)$
so, the equation of required cirde is
$(\text{x}-5)^2+(\text{y}-6)^2=6^2$
$\Rightarrow\text{x}^2+\text{y}^2-10\text{x}-12\text{y}+25=0$

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Question 63 Marks

Find the equation of the circle which circumscribes the triangle formed by the lines
2x + y - 3 = 0, x + y - 1 = 0 and 3x + 2y - 5 = 0

Answer

The given equation of lines
2x + y = 3 ....... (1)
x - y = 1 .......... (2)
3x + 2y = 5 ......... (3)
Let A, B & C are the point of intersection of lines (1) & (2), (2) & (3) and (3) & (1) respectively
$\therefore$ A = (2, -1), B = (3, -2) & C = (1, 1)
Let x² + y²+ 2gx + 2fy + c = 0 ........ (A)
be the circle that circum scnbinq $\Delta\text{ABC}$
$\therefore$ 4 + 1 - 4g - 2f + c = 0 ........ (4)
9 + 16 + 6g + 2f + c = 0 ........ (5)
1 + 1 + 2g + 2f + c = 0 ......... (6)
Solving (4), (5) & (6) we get,
$\text{g}=-\frac{13}{2},\ \text{f}=\frac{5}{2}\ \&\ \text{c}=16$
from (A),
The required cirde is
x² + y² - 13x + 5y - 16 = 0

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Question 73 Marks

Find the equation of a circle,
Which touches both the axes at a distance of 6 units from the origin.

Answer

The circle touches the axes at (0, 6) and (6, 0) respectively
Thus, the centre of circle will be (6, 6) ( as shown in fig)
and radius $=\text{OA}=\sqrt{(6-0)^2+(6-6)^2}=\sqrt{36}=6$ (by distance formulla)
$\therefore$ The equation of circle will be $(\text{x}-6)^2+(\text{y}-6)^2=6^2$
$\Rightarrow\text{x}^2+\text{y}^2-12\text{x}-12\text{y}+36=0$

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Question 83 Marks

Find the equation of the circle having (1, -2) as its centre and passing through the intersection of the lines 3x + y = 14 and 2x + 5y = 18.

Answer

Intersection of 3x + y = 14 and 2x +5y= 18 is
Obtained by solving two equations.
x = 4 and y = 2
Point (4,2) is on circle, hence~ distance from centre
(1,-2)
= Radius
$=\sqrt{(1-4)^2+(-2-1)^2}$
$=\sqrt{9+16}$
$=5$
Equation of the circle with centre (4,2) and radius 5 is,
(x - 1)²+ (y + 2)² = 25
x² - 2x + 1 + y² + 4y + 4 = 25
x² + y² - 2x + 4y - 20 = 0

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Question 93 Marks

If the equations of two diameters of a circle are 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle.

Answer

The equation of two diameters of the circle (x - a)² + (y - b)² = r² ......... (A)
is 2x + y = 6 ......... (1)
3x + 2y = 4 .......... (2)
The point of intersection of (1) & (2) is C = (8, -10), which is the centre of circle.
Also, radius = 10
$\therefore$ (A)
⇒ (x - 8)²+(y + 10)²= 10²
⇒ x² + y²- 16x + 20y + 64 = 0

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Question 103 Marks

Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y − 1 = 0.

Answer

The centre of the required circle in (3,4) and the circle touches the line 5x + 12y = 1
so, radius = OA = Perpendicular distance of O to 5x + 12y = 1
$[\therefore$ radius is perpendicular to the tangent$]$
$\Rightarrow\text{OA}\frac{5\times3+12\times4-1}{\sqrt{5}^2+12^2}$
$=\frac{62}{13}$
Thus the equation of circle will be,
$(\text{x}-3)^2+(\text{y}-4)^2=\Big(\frac{62}{13}\Big)^2$
$\Rightarrow169[\text{x}^2+\text{y}^2-6\text{x}-8\text{y}]+25\times169=3844$
$\Rightarrow169[\text{x}^2+\text{y}^2-6\text{x}-8\text{y}]+381=0$

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Question 113 Marks

Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on the line x - 4y = 1.

Answer

The circle passes through P & Q and the centre lies on
X - 4y = 1 .........(1)
The genral equation of circle is
x² + y² + 2gx + 2fy + c = 0 .......... (2)
$\therefore$ P & Q lies cn (2), so,
9 + 49 + 6g + 14f + c = 0 ............ (3)
25 + 25 + 10g + 10f + c = 0 ......... (4)
Also, centre (-g, -f) lies on (1)
$\therefore$ - g + 4f = 1 .......... (5)
Solving (3) (4) & (5) we get
g = 3, f = 1 & c = -90
from (2)
The equation of cirde is
x² + y²+ 6x + 2y - 90 = 0

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Question 123 Marks

Find the equation of the circle passing through the points:
(5, -8), (-2, 9) and (2, 1)

Answer

We know that the general equation of circle is x² + y² + 2gx + 2fy + c = 0 ........ (1)
We have,
P (5, -8), Q (-2, 9) and R (2, 1)
Since P, Q & R lies on (1)
P, Q & R lies on(1), so,
25 + 64 + 10g + 16f + c = 0 ........ (2)
4 + 81 + 4g - 18f + c = 0 ............ (3)
4 + 1 + 4g - 2f + c = 0 ................ (4)
Solving (2), (3) & (4), we get,
g = 58, f = 24 & c = -285
Thus, equation of eirel e is,
x² + y² + 116x - 48y - 285 = 0 from (1)

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Question 133 Marks

Find the equation of the circle which passes through (3, -2), (-2, 0) and has its centre on the line 2x - y = 3

Answer

A circle pass Ing through P(3, -2) and Q(-2, 0) and having its centre on 2x - y = 3.
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0.
Since the circle passes through (3, -2) andAlso (-2, 0) therefore
9 + 4 + 6g - 4f + c = 0 ......... (1)
4 + 0 - 4g + 0 + c = 0 .......... (2)
Also the centre of the circle lies on 2x - y = 3
-2g + f = 3 ....... (3)
Solving equallons (1), (2) and (3), we get
$\text{g}=\frac{3}{2},\ \text{f}=6$ and $\text{c}=2$
Therefore the equation of the circle Is
x² + y² + 3x + 12y + 2 = 0

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Question 143 Marks

Show that the points (5, 5), (6, 4), (-2, 4) and (7, 1) all lie on a circle, and find its equation, centre and radius.

Answer

The genral equation of circle is
x² + y² + 2gx + 2fy + c = 0 .......... (1)
centre = (-g, -f) and
Rsdius $=\sqrt{\text{g}^2+\text{f}^2-\text{c}}$
$\therefore$ P = (5, 5), Q (6, 4) Be R = (-2, 4) lies on (1)
$\therefore$ 25 + 25 + 10g + 10f + c = 0 ........ (2)
36 + 16 + 12g + 8f + c = 0 ............. (3)
4 + 16 + 4g + 8f + c = 0 ................ (4)
Solving (2) (3) & (4), we get
g = -2, f = -1 & c = -20
from (1)
The equation of cirde is
x² + y²+ 4x + 2y - 20 = 0
Clearly S = (7, 1) Satisfy (A)
Hence P,Q,R,S are concydic
Now
centre = (-g, -f) = (2, 1)
radius $\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{4+1+20}=\sqrt{25}=5$

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Question 153 Marks

Find the equation of the circle passing through the points:
(1, 2), (3, -4) and (5, -6)

Answer

We know that the general equation of circle is x² + y² + 2gx + 2fy + c = 0 ........ (1)
We have,
P (1,2), Q (3, -4) and R (5, -6)
Since P, Q & R lies on (1)
$\therefore$ x² + y² + 2gx + 2fy + c = 0 ...... (1)
1 + 4 + 2g + 4f + c = 0 ................ (2)
9 + 16 + 6g - 8f + c = 0 ............... (3)
25 + 36 + 10g - 12f + c = 0 .......... (4)
Solving (2), (3) & (4), we get,
g = -11, f = -2 & c = 25
from (1)
The equation of circle is
x² + y² -22x - 4y + 25 = 0

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Question 163 Marks

Find the equation of the circle which circumscribes the triangle formed by the lines
x + y + 3 = 0, x - y + 1 = 0 and x = 3

Answer

The given equation of lines
x + y = -3 ......... (1)
x - y = -1 ......... (2)
x = 3 .............. (3)
Let A, B, C are the point of intersection of lines (1) & (2), (2) & (3) and (3) & (1) respectively
$\therefore$ A = (-2, -1), B = (3, 4) & C = (3, -6)
Now A cirde x² + y²+ 2gx + 2fy + c = c ........ (A)
circumscribing the $\Delta\text{ABC}$
$\therefore$ 4 + 1 - 4g - 2f + c = 0 ........ (4)
9 + 16 + 6g + af + c = 0 ........ (5)
9 + 36 + 6g - 12f + c = 0 ....... (6)
Solving (4), (5) & (6) we get,
g = -3, f = 1, c = -15
from (A),
The equation of required cirde is
x² + y² - 6x + 2y - 15 = 0

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Question 173 Marks

Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.

Answer

we have,
P = (3, -2), Q = (1,0), R = (-1,-2) and S = (1, -4)
let us consider A circle x² + y² + 2gx + 2fy + c = 0 ........ (1)
Passes through P, Q & R
$\therefore$ 9 + 4 + 6g - 4f + c = 0 ........ (2)
1 + 0 + 2g - 0 + c = 0 ............ (3)
1 + 4 - 2g - 4f + c = 0 ............ (4)
Solving (2), (3) & (4) we get,
g = -1, f = 2 & c = 1
from(1)
The required equation of ercle is
x² + y² - 2x + 4y + 1 = 0 ......... (5)
Clearly s = (1, -4) satisfy (5)
Thus,
P, Q, R & S are concydic

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Question 183 Marks

If the lines 2x - 3y = 5 and 3x - 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

Answer

Area of given circle is = 154
$\pi^2=154$
$\frac{22}{7}\text{r}^2154$
$\text{r}^2=154\times\frac{7}{22}$
$\text{r}^2=154\times\frac{7}{22}$
$\text{r}^2=49$
$\text{r}=7$
The intersection point of 2x - 3y = 5 and 3x - 4y = 7 is
The centre of the circle.
Solving simultaneous equations
2x - 3y = 5 and 3x - 4y = 7 we get,
Centre of circle as (1, -1)
Equation of circle with centre (1, -1) and radius= 7 is,
(x - 1)² + (y + 1)² = 72
x² - 2x + 1 + y² + 2y + 1= 49
x² - 2x + y² + 2y = 47

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3 Marks Question - MATHS STD 11 Science Questions - Vidyadip