# SSC > Circles

Explore popular questions from Circles for SSC. This collection covers Circles previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

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Incorrect-0.5

The square of the length of the tangent from {tex} \left(3,\ -4\right) {/tex} to the circle {tex} x^{2}+y^{2}-4x-6y+3=0 {/tex} is:

A

20

B

30

40

D

50

##### Explanation

Length of tangent from the point {tex} \left(x_{1},y_{1}\right) {/tex} to the circle {tex} x^{2} + y^{2} +2gx + 2fy + c = 0 {/tex} is

{tex} \sqrt{x_{1}^{2} +y_{1}^{2} + 2gx_{1} + 2fy_{1} + c } {/tex}
Required length of tangent from the point (3,-4) to the circle {tex} x^{2} + y^{2} -4x -6y + 3 = 0 {/tex}

{tex} \sqrt{3^{2} + 4^{2} + -4(3) - 6(-4) + 3} = \sqrt{40}{/tex}
Square of the length of tangenet is 40

Q 2.

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If {tex} g^{2}+f^{2}=c {/tex} then the equation {tex} x^{2}+y^{2}+2gx+2fy+c=0 {/tex} will represent:

A

B

C

a circle of diameter {tex} \sqrt{c} {/tex}

##### Explanation

Given that {tex} x^{2}+y^{2}+2gx+2fy+c=0 {/tex}

and {tex} g^{2}+f^{2}=c {/tex}

Radius of circle = {tex} \sqrt{g^{2}+f^{2}-c} {/tex}

{tex} \therefore g^{2}+f^{2}=c {/tex}

Thus given equation represents a circle of radius 0.

Q 3.

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The limit of the perimeter of the regular n polygons inscribe in a circle of radius R as {tex} n\ \rightarrow\ \infty {/tex} is:

{tex} 2 \pi R {/tex}

B

{tex} \pi R {/tex}

C

{tex} 4R {/tex}

D

{tex} \pi R^{2} {/tex}

##### Explanation

As {tex} n\ \rightarrow\ \infty {/tex}, therefore polygon becomes a circle and perimeter of circle = {tex} 2 \pi R {/tex}.

Q 4.

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The value of n, for which the circle {tex} x^{2}+y^{2}+2nx+6y+1=0 {/tex} intersects the circle {tex} x^{2}+y^{2}+4x+2y=0 {/tex} orthogonally is:

A

{tex} \frac{11}{8} {/tex}

B

-1

{tex} \frac{-5}{4} {/tex}

D

{tex} \frac{5}{2} {/tex}

##### Explanation

Equations of circles are

{tex} x^{2}+y^{2}+2nx+6y+1=0 {/tex}

and {tex} x^{2}+y^{2}+4x+2y=0 {/tex}

Since, the circles cut orthogonally

{tex} \therefore\ 2gg+2ff = c+c {/tex}

=> {tex} 2n \times 2+6 \times 1 = 1+0 {/tex}

=> {tex} 4h+6 = 1 {/tex}

=> {tex} n = \frac{-5}{4} {/tex}

Q 5.

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The Value of c for which the line {tex} y=2x+c {/tex} is a tangent to the circle {tex} x^{2}+y^{2}=16 {/tex} is:

A

{tex} -16\sqrt{5} {/tex}

{tex} 4\sqrt{5} {/tex}

C

{tex} 16\sqrt{5} {/tex}

D

20

##### Explanation

Given that {tex} y = 2x+c {/tex} ...(i)

{tex} x^{2}+y^{2}=16 {/tex} ...(ii)

We Know that, if {tex} y = mx+c {/tex} is tangent to the circle {tex} x^{2}+y^{2}=a^{2} {/tex} then {tex} c = \pm a\sqrt{1+m^{2}} {/tex} here m = 2, a = 4

{tex} c = \pm 4\sqrt{1+2^{2}} = \pm 4\sqrt{5} {/tex}

Q 6.

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The radical axis of two circle and line joining their centres are:

A

Parallel

Perpendicular

C

Neither Parallel nor perpendicular

D

Intersecting but not perpendicular

##### Explanation

Radical axis is the common chord of the two circles and radical axis is perpendicular to the line joining the centers of two circles.

Q 7.

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Which of the following is a point on the common chord of the circles {tex} x^{2}+y^{2}+2x-3y+6=0\ and\ x^{2}+y^{2}+x-8y-13=0 {/tex}?

A

{tex} \left(1,\ -2 \right) {/tex}

B

{tex} \left(1,\ 4\right) {/tex}

C

{tex} 1,\ 2 {/tex}

{tex} \left(1,\ -4\right) {/tex}

##### Explanation

Let the equation of circles are

{tex} S_{1} = x^{2}+y^{2}+2x-3y+6=0 {/tex} ...(i)

{tex} S_{2} = x^{2}+y^{2}+x-8y-13=0 {/tex} ...(ii)

Equation of common chord is

{tex} S_{1} - S_{2} = 0 {/tex}

=> {tex} \left(x^{2}+y^{2}+2x-3y+6\right) - \left(x^{2}+y^{2}+x-8y-13\right) = 0 {/tex}

=> {tex} x+5y+19=0 {/tex} ...(iii)

In the given options only the point {tex} \left(1,\ -4\right) {/tex} satisfied the eq. (iii)

Q 8.

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The radius of the circle passing through the point {tex} \left(6,\ 2\right) {/tex} and two of whose diameter are {tex} x+y=6\ and\ x+2y=4 {/tex} is:

A

4

B

6

C

20

{tex} \sqrt{20} {/tex}

##### Explanation

Centre is the point of intersection of two diameters i.e., the point of intersection of two diameter is {tex} C \left(8,\ -2\right) {/tex} therefore the distance from the centre to the point {tex} P \left(6,\ 2\right) {/tex} is

{tex} r = CP = \sqrt{4+16} = \sqrt{20} {/tex}

Q 9.

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The radius of any circle touching the lines {tex} 3x-4y+5=0\ and\ 6x-8y-9=0 {/tex} is:

A

1.9

0.95

C

2.9

D

1.45

##### Explanation

Since, given lines are parallel to each other so the line segment joining the points of contact is diameter of the circle, distance between the lines

{tex} 3x-4y+5=0\ and\ 3x-4y-\frac{9}{2}=0\ is {/tex}

{tex} |\frac{5+\frac{9}{2}}{\sqrt{3^{2}+4^{2}}}|\ =\ |\frac{9}{10}| = 1.9 {/tex}

Length of diameter of the circle is 1.9.

Radius of circle = {tex} \frac{1.9}{2} = 0.95 {/tex}

Q 10.

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The locus of the middle point of the chords of the circle {tex} x^{2}+y^{2}=a^{2} {/tex} such that the chords pass through a given point {tex} \left(x_{1},\ y_{1}\right) {/tex} is:

{tex} x^{2}+y^{2}-xx_{1}-yy_{1}=0 {/tex}

B

{tex} x^{2}+y^{2}=x^{2}_{1}+y^{2}_{1} {/tex}

C

{tex} x+y=x_{1}+y'_{2} {/tex}

D

{tex} x+y=x^{2}_{1}+y^{2}_{1} {/tex}

##### Explanation

Let {tex} P \left(x_{1}, y_{1}\right) {/tex} be the point, then the chord of contact of tangents drawn from P to the circle

{tex} x^{2}+y^{2}=a^{2}\ is\ xx_{1}+yy_{1}=a^{2} {/tex}

{tex} \therefore\ x^{2}+y^{2} = a^{2} \left(\frac{xx_{1}+yy_{1}}{a^{2}}\right) {/tex}

=> {tex} x^{2}+y^{2}-xx_{1}-yy_{1}=0 {/tex}

Q 11.

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The equations of the tangents to the circle {tex} x^{2}+y^{2}-6x+4y-12=0 {/tex} which are parallel to the line {tex} 4x+3y+5=0 {/tex},are

A

{tex} 4x+3y+11=0 {/tex} and {tex} 4x+3y+8=0 {/tex}

B

{tex} 4x+3y-9=0 {/tex} and {tex} 4x+3y+7=0 {/tex}

{tex} 4x+3y+19=0 {/tex} and {tex} 4x+3y-31=0 {/tex}

D

{tex} 4x+3y-10=0 {/tex} and {tex} 4x+3y+12=0 {/tex}

##### Explanation

The centre and radius of given circle are {tex} \left(3,\ -2\right) {/tex} and 5 respectively. The equation of a line parallel to {tex} 4x+3y+5=0\ is\ 4x+3y+n=0 {/tex}

As we know that perpendicular distance from centre {tex} \left(3,\ -2\right) {/tex} to the circle = radius of the circle.

{tex} |\frac{4 \times 3+3 \times \left(-2\right)+n }{\sqrt{4^{2}+3^{2}}}| {/tex}

{tex} n = 19,\ -31 {/tex} => Equation of tangents are

{tex} 4x+3y+19=0\ and\ 4x+3y-31=0 {/tex}

Q 12.

Correct2

Incorrect-0.5

The lines {tex} 2x-3y=5\ and\ 3x-4y=7 {/tex} are diameters of a circle having area as 154 sq unit. Then the equation of the circle is:

A

{tex} x^{2}+y^{2}+2x-2y=62 {/tex}

B

{tex} x^{2}+y^{2}+2x-2y=47 {/tex}

{tex} x^{2}+y^{2}-2x+2y=47 {/tex}

D

{tex} x^{2}+y^{2}-2x+2y=62 {/tex}

##### Explanation

The equation of diameters are

{tex} 2x-3y = 5 {/tex} ...(i)

and {tex} 3x-4y = 7 {/tex} ...(ii)

On solving eqs. (i) and (ii) we get

{tex} x = 1\ and\ y = -1 {/tex}

Therefore, Centre of circle = {tex} \left(1,\ -1\right) {/tex}

Since area of circle is 154 sq unit then radius of circle r = 7

Equation of circle is {tex} \left(x-1\right)^{2}+ \left(y+1\right)^{2} = 49 {/tex}

=> {tex} x^{2}+y^{2}-2x+2y = 47 {/tex}

Q 13.

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If the circle {tex} x^{2}+y^{2}+6x-2y+k=0 {/tex} bisects the circumference of the circle {tex} x^{2}+y^{2}+2x-6y-15=0 {/tex} then k is equal to:

A

21

B

-21

C

23

-23

##### Explanation

The condition for a circle bisecting the circumference of the second circle is:

{tex} 2g_{2} \left(g_{1}-g_{2}\right)+2f_{2} \left(f_{1}-f_{2}\right)=c_{1}-c_{2} {/tex}

=> {tex} 2 \left(1\right) \left(3-1\right)+2 \left(-3\right) \left(-1+3\right)=k+15 {/tex}

=> {tex} -8 = k + 15 => k = -23 {/tex}

Q 14.

Correct2

Incorrect-0.5

If {tex} 5x-12y=10 {/tex} and {tex} 12y-5x+16=0 {/tex} are two tangents to a circle, then the radius of the circle is

1

B

2

C

4

D

6

##### Explanation

Given tangents are {tex} 5x-12y+10=0, {/tex}

and {tex} 5x-12y-16=0 {/tex} parallel.

Radius = {tex} \frac{c_{1} - c_{2}}{2\sqrt{a^{2} + b^{2}}} {/tex}

Radius = {tex} \frac{c_{1}-c_{2}}{2\sqrt{5^{2}+ \left(-12\right)^{2} }} = \frac{26}{2 \times 13} = 1 {/tex}

Q 15.

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Incorrect-0.5

Equation of the circle passing through the point {tex} \left(3,\ 4\right) {/tex} and concentric with the circle {tex} x^{2}+y^{2}-2x-4y+1=0 {/tex} is

A

{tex} x^{2}+y^{2}-2x-4y=0 {/tex}

B

{tex} x^{2}+y^{2}-2x-4y+3=0 {/tex}

{tex} x^{2}+y^{2}-2x-4y-3=0 {/tex}

D

none of the above

##### Explanation

Let the equation of the concentric circle {tex} b\ x^{2}+y^{2}-2x-4y+n=0 {/tex} it passes through {tex} \left(3,\ 4\right) {/tex}

{tex} 3^{2}+4^{2}-2 \left(3\right)-4 \left(4\right)+n=0 {/tex}

n = -3

Thus the equation of concentric circle is

{tex} x^{2}+y^{2}-2x-4y-3=0 {/tex}

Q 16.

Correct2

Incorrect-0.5

A line through {tex} P \left(1,\ 4\right) {/tex} intersect a circle {tex} x^{2}+y^{2}=16 {/tex} at A and B, then PA-PB is equal to:

1

B

2

C

3

D

4

##### Explanation

We know {tex} PA,\ PB = PT^{2} {/tex} (PT is length of tangent)

Let {tex} S = x^{2}+y^{2}-16 {/tex}

{tex} PT = \sqrt{S_{1}} {/tex}

= {tex} \sqrt{1+16-16} = 1 {/tex}

Q 17.

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Incorrect-0.5

The number of common tangents to the circles {tex} x^{2}+y^{2}-2x-4y+1=0 {/tex} and {tex} x^{2}+y^{2}-12x-16y+91=0 {/tex} is

A

1

B

2

C

3

4

Q 18.

Correct2

Incorrect-0.5

A, B, C and D are the points of intersection with the co-ordinate axes of the lines {tex} ax+by=ab\ and\ bx+ay=ab {/tex} then:

A, B, C, D are concyclic

B

A, B, C, D form a parallelogram

C

A, B, C, D form a rhombus

D

none of the above

##### Explanation

Here {tex} OA.OB = OC.OD {/tex}

So, point A, B, C and D are concylic

Q 19.

Correct2

Incorrect-0.5

The gradient of the radical axis of the circles {tex} x^{2}+y^{2}-3x-4y+5=0\ and\ 3x^{2}+3y^{2}-7x+8y+11=0 {/tex} is

A

{tex} \frac{1}{3} {/tex}

{tex} -\frac{1}{10} {/tex}

C

{tex} -\frac{1}{2} {/tex}

D

{tex} -\frac{2}{3} {/tex}

##### Explanation

Let the equation of circles be

{tex} S_{1} = x^{2}+y^{2}-3x-4y+5=0 {/tex}

and {tex} S_{2} = x^{2}+y^{2}-\frac{7}{3}x+\frac{8}{3}y+\frac{11}{3}=0 {/tex}

The equation of radical axis is {tex} S_{1} - s_{2} = 0 {/tex}

=> {tex} -2x-20y-4=0 {/tex}

=> {tex} y = -\frac{x}{10}-\frac{1}{5} {/tex}

Q 20.

Correct2

Incorrect-0.5

The limiting point of the system of circles represented by the equation {tex} 2\left(x^{2}+y^{2}\right)+nx+\frac{9}{2}=0 {/tex} are

{tex} \left(\pm \frac{3}{2},\ 0\right) {/tex}

B

{tex} \left(0,\ 0\right)\ and\ \left(\frac{9}{2},\ 0\right) {/tex}

C

{tex} \left(\pm \frac{9}{2},\ 0\right) {/tex}

D

{tex} \left(\pm 3,\ 0\right) {/tex}

##### Explanation

We have {tex} 2 \left(x^{2}+y^{2}\right)+nx+\frac{9}{2}=0 {/tex}

On comparing with standard equation of circle,

We get centre {tex} \left(-\frac{n}{4},\ 0\right) {/tex} and radius {tex} r = \sqrt{\frac{n^{2}}{16}-\frac{9}{4}} {/tex}

We know that for limiting point

{tex} \frac{n^{2}}{16}-\frac{9}{4}=0 {/tex}

=> {tex} n^{2} = 36 {/tex}

=> {tex} n = \pm 6 {/tex}

Limiting points are {tex} \left(\pm \frac{6}{4},\ 0\right)\ or\ \left(\pm \frac{3}{2},\ 0\right) {/tex}

Q 21.

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The radical centre of the circles {tex} x^{2}+y^{2}-16x+60=0,{/tex} {tex} x^{2}+y^{2}-12x+27=0,{/tex} {tex} x^{2}+y^{2}-12y+8=0 {/tex} is

A

{tex} \left(13,\ \frac{33}{4}\right) {/tex}

B

{tex} \left(\frac{33}{4},\ -13\right) {/tex}

C

{tex} \left(\frac{33}{4},\ 13\right) {/tex}

none of these

##### Explanation

Let the equation of circles are

{tex} S_{1} = x^{2}+y^{2}-16x+60=0 {/tex} ...(i)

{tex} S_{2} = x^{2}+y^{2}-12x+27=0 {/tex} ...(ii)

and {tex} S_{3} = x^{2}+y^{2}-12y+8=0 {/tex} ...(iii)

The radical axis of circles (i) and (ii) is

{tex} S_{1} - S_{2} = 0 {/tex}

=> {tex} \left(x^{2}+y^{2}-16x+60\right) - \left(x^{2}+y^{2}-12x+27\right) = 0 {/tex}

=> {tex} -4x+33 = 0 {/tex} ...(iv)

The radical axis of circles (ii) and (iii) is

{tex} S_{2} - S_{3} = 0 {/tex}

=> {tex} x^{2}+y^{2}-12x+27- \left(x^{2}+y^{2}-12y+8\right)=0 {/tex}

{tex} -12x+12y+19=0 {/tex} ...(v)

Solving equations (iv) and (v) we get radical centre {tex} \left(\frac{33}{4},\ \frac{20}{3}\right) {/tex}

Q 22.

Correct2

Incorrect-0.5

The circles {tex} x^{2} + y^{2} - 10x +16 = 0 {/tex} and {tex} x^{2} + y^{2} = r^{2} {/tex} intersect each other at two distinct points if:

A

r < 2

B

r > 8

2 < r < 8

D

{tex} 2 \le r \le 8 {/tex}

##### Explanation

The equation of the given circles are

{tex} x^{2}+y^{2}-10x+16=0 {/tex}

=> {tex} \left(x-5\right)^{2}+y^{2}=3^{2} {/tex}

Whose centre is {tex} \left(5,\ 0\right) {/tex} and radius = 3

and {tex} x^{2}+y^{2}=r^{2} {/tex}

Whose centre is {tex} \left(0,\ 0\right) {/tex} and radius = r

Clearly, these two circle will intersect each other at two distinct point if {tex} r > OA {/tex}

=> {tex} r > 5-3 => r > 2\ and\ r < OB {/tex}

=> {tex} r < 2+3+3 {/tex}

=> {tex} r < 8 {/tex}

=> {tex} 2 < r < 8 {/tex}

Q 23.

Correct2

Incorrect-0.5

The centres of a set of circles, each of radius 3, lies on the circle {tex} x^{2} + y^{2} = 25 {/tex}. The locus of any point in the set is:

{tex} 4 \le x^{2}+y^{2} \le 64 {/tex}

B

{tex} x^{2}+y^{2} \le 25 {/tex}

C

{tex} x^{2}+y^{2}\ge 25 {/tex}

D

{tex} 3 \le x^{2}+y^{2} \le 9 {/tex}

##### Explanation

Let {tex} \left(h,\ k\right) {/tex} be the centre of a circle, then equation of circle.

{tex} \left(x-h \right)^{2}+ \left(y-k \right)^{2}=9 {/tex}

This centre lies on {tex} x^{2}+y^{2}-25 {/tex}

=> {tex} h^{2}+k^{2}=25 {/tex}

{tex} \therefore 2 \le {/tex} distance between the centre of the two circles {tex} \le 8 {/tex}.

=> {tex} 2 \le \sqrt{h^{2}+k^{2}} \le 8 {/tex}

=> {tex} 4 \le \left(h^{2}+k^{2}\right) \le 64 {/tex}

Therefore, Locus of {tex} \left(h,\ k\right)\ is\ 4 \le \left(x^{2}+y^{2}\right) \le 64 {/tex}

Q 24.

Correct2

Incorrect-0.5

A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is:

{tex} \left(x-p\right)^{2}=4qy {/tex}

B

{tex} \left(x-q\right)^{2}=4py {/tex}

C

{tex} \left(y-p\right)^{2}=4qx {/tex}

D

{tex} \left(y-q\right)^{2}=4py {/tex}

##### Explanation

In a circle AB is as a diameter where the coordinate of A is {tex} \left(p,\ q \right) {/tex} and let the co-ordinate at B is {tex} \left(x_{1},\ y_{1}\right) {/tex}.

Equation of circle in diameter from is

{tex} \left(x-p\right) \left(x-x_{1}\right)+ \left(y-q\right) \left(y-y_{1}\right)=0 {/tex}

{tex} x^{2}- \left(p+x_{1}\right)x+px_{1}+y^{2}{/tex} {tex} - \left(y_{1}+q\right)y+qy_{1}=0 {/tex}

{tex} x^{2}- \left(p+x_{1}\right)x+y^{2}- {/tex} {tex} \left(y_{1}+q\right)y+px_{1}+qy_{1}=0 {/tex}

Since, the circle touches x-axis.

{tex} \therefore y =0 {/tex}

=> {tex} x^{2}- \left(p+x_{1}\right)x+px_{1}+qy_{1}=0 {/tex}

Also, the discriminant of above equation will be equal to zero because circle touches x-axis

{tex} \therefore \left(p+x_{1}\right)^{2}= 4 \left(px_{1}+qy_{1} \right){/tex} {tex} => p^{2}+x^{2}_{1}+2px_{1}=4px_{1}+4qy_{1} {/tex}

=> {tex} x^{2}_{1}-2px_{1}+p^{2}=4qy_{1} {/tex}

Therefore the locus of point B is {tex} \left(x-p\right)^{2}=4qy {/tex}

Q 25.

Correct2

Incorrect-0.5

If the lines {tex} 2x + 3y + 1 = 0 {/tex} and {tex} 3x - y - 4 = 0 {/tex} lies along diameter of a circle of circumference {tex} 10 \pi {/tex}, then the equation of the circle is:

{tex} x^{2} + y^{2} - 2x + 2y - 23 = 0 {/tex}

B

{tex} x^{2} + y^{2} - 2x - 2y - 23 = 0 {/tex}

C

{tex} x^{2} + y^{2} + 2x + 2y - 23 = 0 {/tex}

D

{tex} x^{2} + y^{2} + 2x - 2y - 23 = 0 {/tex}

##### Explanation

The lines {tex} 2x+3y+1=0\ and\ 3x-y-4=0 {/tex} are diameter of circle

On solving these equations, we get

{tex} x=1\ and\ y=-1 {/tex}

Therefore the centre or circle = {tex} = \left(1-1\right) {/tex}) and

Circumference = 10{tex} \pi {/tex}

=> {tex} 2 \pi r = 10 \pi => r = 5 {/tex}

Equation of circle

{tex} \left(x-x_{1}\right)^{2}+ \left(y-y_{1}\right)^{2}=r^{2} {/tex}

=> {tex} \left(x-1\right)^{2}+ \left(y+1\right)^{2}=5^{2} {/tex}

=> {tex} x^{2}+1-2x+y^{2}+2y+1=25 {/tex}

=> {tex} x^{2}+y^{2}-2x+2y-23=0 {/tex}