JEE MainMathematics

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Circles: JEE Main Mathematics Question with Solution

Q: Why is the radius of the new circle equal to the perpendicular distance from C to AB?

A circle with centre has the line as a tangent. The radius drawn to the point of tangency is perpendicular to the tangent line. Therefore, the radius equals the perpendicular distance from the centre to the tangent line .

Let the tangents at the points

A(4,-11)

and

B(8,-5)

on the circle

x^2+y^2-3 x+10 y-15=0

, intersect at the point

C

. Then the radius of the circle, whose centre is

C

and the line joining

A

and

B

is its tangent, is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

2 \sqrt{13}

\sqrt{13}

\frac{3 \sqrt{3}}{4}

\frac{2 \sqrt{13}}{3}

Step-by-step solutionView

Correct answer

The radius of the circle centered at $C$ tangent to chord $AB$ is $\frac{2\sqrt{13}}{3}$ .

Option analysis

Why each option works or fails

A ·

2 \sqrt{13}

Calculates the half-length of chord $AB$ ( $AM = \sqrt{13}$ ) and multiplies by 2 to give the whole chord length instead of finding the distance $CM$ . The required radius is the perpendicular distance from $C$ to chord $AB$ , which is $CM$ , not the length of chord $AB$ .

B ·

\sqrt{13}

Confuses the required radius (the segment $CM$ ) with the half-length of the chord $AM$ . Recognize that $AM$ lies along chord $AB$ , whereas the distance from center $C$ to chord $AB$ is the segment $CM = \frac{AM^2}{OM}$ .

C ·

\frac{3 \sqrt{3}}{4}

Makes an arithmetic slip while squaring coordinates or evaluating the radius of the original circle. Carefully compute the center $\left(\frac{3}{2}, -5\right)$ and radius squared $R^2 = \frac{9}{4} + 25 + 15 = \frac{169}{4}$ , giving $R = \frac{13}{2}$ .

D ·

\frac{2 \sqrt{13}}{3}

None. The student correctly applies geometric properties of the right-angled triangle formed by the circle center, tangent point, and tangent intersection. This is the correct value: $CM = \frac{AM^2}{OM} = \frac{13}{39/4} = \frac{52}{39} = \frac{4}{3}$ , leading to radius $r = CM = \frac{2\sqrt{13}}{3}$ .

Reviewed route

Solution

StepWorking

01given

Circle $S \equiv x^2+y^2-3x+10y-15=0$ , points on circle $A(4,-11)$ and $B(8,-5)$ . Tangents at $A$ and $B$ meet at $C(h,k)$ .

02goal

Find the radius of the circle with centre $C$ that touches the chord $AB$ . This radius is simply the perpendicular distance from $C$ to line $AB$ .

03approach

1. Find the equation of chord $AB$ using two-point form. 2. Since $AB$ is the chord of contact from $C(h,k)$ , write $T=0$ and compare coefficients with the line equation of $AB$ to determine $(h,k)$ . 3. Calculate the perpendicular distance from $(h,k)$ to line $AB$ .

04execute

Slope of line $AB = \frac{-5 - (-11)}{8 - 4} = \frac{6}{4} = \frac{3}{2}$ . Equation of line $AB$ : $y + 5 = \frac{3}{2}(x - 8) \implies 3x - 2y - 34 = 0$ .

05execute

Chord of contact from $C(h,k)$ is $T = 0$ : $hx + ky - \frac{3}{2}(x+h) + 5(y+k) - 15 = 0$ $\implies \left(h - \frac{3}{2}\right)x + (k + 5)y - \left(\frac{3}{2}h - 5k + 15\right) = 0$ . Comparing with $3x - 2y - 34 = 0$ : $\frac{h - 3/2}{3} = \frac{k + 5}{-2} = \frac{\frac{3}{2}h - 5k + 15}{34}$ . Let this ratio be $\lambda$ . Then $h - \frac{3}{2} = 3\lambda \implies h = 3\lambda + \frac{3}{2}$ , and $k = -2\lambda - 5$ . Substitute into third term: $\frac{3}{2}\left(3\lambda + \frac{3}{2}\right) - 5(-2\lambda - 5) + 15 = 34\lambda$ $\implies \frac{9}{2}\lambda + \frac{9}{4} + 10\lambda + 25 + 15 = 34\lambda$ $\implies \frac{29}{2}\lambda + \frac{169}{4} = 34\lambda \implies \frac{39}{2}\lambda = \frac{169}{4} \implies \lambda = \frac{13}{6}$ . Then $h = 3\left(\frac{13}{6}\right) + \frac{3}{2} = \frac{13}{2} + \frac{3}{2} = 8$ , and $k = -2\left(\frac{13}{6}\right) - 5 = -\frac{13}{3} - 5 = -\frac{28}{3}$ .

06execute

The required radius $r$ is the distance from $C\left(8, -\frac{28}{3}\right)$ to line $3x - 2y - 34 = 0$ : $r = \frac{|3(8) - 2(-28/3) - 34|}{\sqrt{3^2 + (-2)^2}} = \frac{|24 + 56/3 - 34|}{\sqrt{13}} = \frac{|-10 + 56/3|}{\sqrt{13}} = \frac{26/3}{\sqrt{13}} = \frac{2\sqrt{13}}{3}$ .

✓verify

Length of chord $AB = \sqrt{(8-4)^2 + (-5 - (-11))^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$ . Half chord length is $M A = \sqrt{13}$ . Center of original circle is $O(3/2, -5)$ , radius $R = \sqrt{9/4 + 25 + 15} = \sqrt{169/4} = 13/2 = 6.5$ . Distance from $O$ to line $AB$ is $d = \frac{|3(3/2) - 2(-5) - 34|}{\sqrt{13}} = \frac{|9/2 - 24|}{\sqrt{13}} = \frac{39/2}{\sqrt{13}} = \frac{3\sqrt{13}}{2}$ . Check: $d^2 + (AB/2)^2 = \frac{117}{4} + 13 = \frac{169}{4} = R^2$ , correct. In right triangle $OAC$ , $AM \perp OC$ , so $CM = \frac{AM^2}{OM} = \frac{13}{3\sqrt{13}/2} = \frac{26}{3\sqrt{13}} = \frac{2\sqrt{13}}{3}$ . This perfectly matches $r$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is the radius of the new circle equal to the perpendicular distance from C to AB?

A circle with centre $C$ has the line $AB$ as a tangent. The radius drawn to the point of tangency is perpendicular to the tangent line. Therefore, the radius equals the perpendicular distance from the centre $C$ to the tangent line $AB$ .

Why does OC perpendicularly bisect AB?

Because tangents from an external point C to a circle are equal ( $CA = CB$ ) and radii are equal ( $OA = OB$ ), so $OC$ is the axis of symmetry of the kite $OACB$ , meaning $OC \perp AB$ and bisects it.

Answer

The radius of the circle centered at $C$ tangent to chord $AB$ is $\frac{2\sqrt{13}}{3}$ .

Why each option works or fails

A: $2 \sqrt{13}$ - Calculates the half-length of chord $AB$ ( $AM = \sqrt{13}$ ) and multiplies by 2 to give the whole chord length instead of finding the distance $CM$ . The required radius is the perpendicular distance from $C$ to chord $AB$ , which is $CM$ , not the length of chord $AB$ .
B: $\sqrt{13}$ - Confuses the required radius (the segment $CM$ ) with the half-length of the chord $AM$ . Recognize that $AM$ lies along chord $AB$ , whereas the distance from center $C$ to chord $AB$ is the segment $CM = \frac{AM^2}{OM}$ .
C: $\frac{3 \sqrt{3}}{4}$ - Makes an arithmetic slip while squaring coordinates or evaluating the radius of the original circle. Carefully compute the center $\left(\frac{3}{2}, -5\right)$ and radius squared $R^2 = \frac{9}{4} + 25 + 15 = \frac{169}{4}$ , giving $R = \frac{13}{2}$ .
D · correct: $\frac{2 \sqrt{13}}{3}$ - None. The student correctly applies geometric properties of the right-angled triangle formed by the circle center, tangent point, and tangent intersection. This is the correct value: $CM = \frac{AM^2}{OM} = \frac{13}{39/4} = \frac{52}{39} = \frac{4}{3}$ , leading to radius $r = CM = \frac{2\sqrt{13}}{3}$ .

Step-by-step solution

given: Circle $S \equiv x^2+y^2-3x+10y-15=0$ , points on circle $A(4,-11)$ and $B(8,-5)$ . Tangents at $A$ and $B$ meet at $C(h,k)$ .
goal: Find the radius of the circle with centre $C$ that touches the chord $AB$ . This radius is simply the perpendicular distance from $C$ to line $AB$ .
approach: 1. Find the equation of chord $AB$ using two-point form. 2. Since $AB$ is the chord of contact from $C(h,k)$ , write $T=0$ and compare coefficients with the line equation of $AB$ to determine $(h,k)$ . 3. Calculate the perpendicular distance from $(h,k)$ to line $AB$ .
execute: Slope of line $AB = \frac{-5 - (-11)}{8 - 4} = \frac{6}{4} = \frac{3}{2}$ . Equation of line $AB$ : $y + 5 = \frac{3}{2}(x - 8) \implies 3x - 2y - 34 = 0$ .
execute: Chord of contact from $C(h,k)$ is $T = 0$ : $hx + ky - \frac{3}{2}(x+h) + 5(y+k) - 15 = 0$ $\implies \left(h - \frac{3}{2}\right)x + (k + 5)y - \left(\frac{3}{2}h - 5k + 15\right) = 0$ . Comparing with $3x - 2y - 34 = 0$ : $\frac{h - 3/2}{3} = \frac{k + 5}{-2} = \frac{\frac{3}{2}h - 5k + 15}{34}$ . Let this ratio be $\lambda$ . Then $h - \frac{3}{2} = 3\lambda \implies h = 3\lambda + \frac{3}{2}$ , and $k = -2\lambda - 5$ . Substitute into third term: $\frac{3}{2}\left(3\lambda + \frac{3}{2}\right) - 5(-2\lambda - 5) + 15 = 34\lambda$ $\implies \frac{9}{2}\lambda + \frac{9}{4} + 10\lambda + 25 + 15 = 34\lambda$ $\implies \frac{29}{2}\lambda + \frac{169}{4} = 34\lambda \implies \frac{39}{2}\lambda = \frac{169}{4} \implies \lambda = \frac{13}{6}$ . Then $h = 3\left(\frac{13}{6}\right) + \frac{3}{2} = \frac{13}{2} + \frac{3}{2} = 8$ , and $k = -2\left(\frac{13}{6}\right) - 5 = -\frac{13}{3} - 5 = -\frac{28}{3}$ .
execute: The required radius $r$ is the distance from $C\left(8, -\frac{28}{3}\right)$ to line $3x - 2y - 34 = 0$ : $r = \frac{|3(8) - 2(-28/3) - 34|}{\sqrt{3^2 + (-2)^2}} = \frac{|24 + 56/3 - 34|}{\sqrt{13}} = \frac{|-10 + 56/3|}{\sqrt{13}} = \frac{26/3}{\sqrt{13}} = \frac{2\sqrt{13}}{3}$ .
verify: Length of chord $AB = \sqrt{(8-4)^2 + (-5 - (-11))^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$ . Half chord length is $M A = \sqrt{13}$ . Center of original circle is $O(3/2, -5)$ , radius $R = \sqrt{9/4 + 25 + 15} = \sqrt{169/4} = 13/2 = 6.5$ . Distance from $O$ to line $AB$ is $d = \frac{|3(3/2) - 2(-5) - 34|}{\sqrt{13}} = \frac{|9/2 - 24|}{\sqrt{13}} = \frac{39/2}{\sqrt{13}} = \frac{3\sqrt{13}}{2}$ . Check: $d^2 + (AB/2)^2 = \frac{117}{4} + 13 = \frac{169}{4} = R^2$ , correct. In right triangle $OAC$ , $AM \perp OC$ , so $CM = \frac{AM^2}{OM} = \frac{13}{3\sqrt{13}/2} = \frac{26}{3\sqrt{13}} = \frac{2\sqrt{13}}{3}$ . This perfectly matches $r$ .

Shortcut:

When to use it: Fastest in geometry: use right-angled triangle $OAC$ with $AM \perp OC$ directly without finding coordinates of $C$ .

given: Circle has centre $O(3/2, -5)$ and radius $R = \sqrt{(3/2)^2 + (-5)^2 - (-15)} = \frac{13}{2}$ . Points $A(4,-11)$ and $B(8,-5)$ lie on the circle.

goal: Find $CM$ , the altitude from $C$ to the chord $AB$ , which equals the perpendicular distance from $C$ to line $AB$ .

approach: In quadrilateral $OACB$ , $OA \perp AC$ (tangent theorem). $OC$ is the perpendicular bisector of chord $AB$ , intersecting $AB$ at midpoint $M$ . In right $\triangle OAC$ , $AM$ is the altitude to hypotenuse $OC$ . By similar triangles: $AM^2 = OM \cdot CM \implies CM = \frac{AM^2}{OM}$ .

execute: Distance $AB = \sqrt{(8-4)^2 + (-5 - (-11))^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$ . Thus $AM = \frac{AB}{2} = \sqrt{13}$ , and $AM^2 = 13$ . In right $\triangle OAM$ , $OM = \sqrt{OA^2 - AM^2} = \sqrt{R^2 - AM^2} = \sqrt{\frac{169}{4} - 13} = \sqrt{\frac{117}{4}} = \frac{3\sqrt{13}}{2}$ . Therefore, $r = CM = \frac{AM^2}{OM} = \frac{13}{\frac{3\sqrt{13}}{2}} = \frac{26}{3\sqrt{13}} = \frac{2\sqrt{13}}{3}$ .

verify: Check dimensions and ratio: $OM \cdot CM = \frac{3\sqrt{13}}{2} \cdot \frac{2\sqrt{13}}{3} = 13 = AM^2$ . Correct.

Circles: JEE Main Mathematics Question with Solution

What feels right?

The radius of the circle centered at $C$ tangent to chord $AB$ is $\frac{2\sqrt{13}}{3}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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