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SSC > Circles

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

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

a Circle of radius g

B

a circle of radius f

C

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

a circle of radius 0

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}

=> Radius = 0

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.

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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.

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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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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.

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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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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.

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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.

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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}

Gradient of radical axis = {tex} -\frac{1}{10} {/tex}

Q 20.

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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.

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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.

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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.

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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.

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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}