Why does represent the circle?

Any point on a circle with diameter endpoints and subtends a right angle at the diameter, which gives via dot product of perpendicular vectors.

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Circles: JEE Main Mathematics Question with Solution

Q: Why can we use the chord of contact formula here?

Since is the intersection of the tangents at and , the line joining the points of contact and is by definition the chord of contact of point .

Points

\text{P}(-3,2), \text{Q}(9,10)

and

(\text{a},4)

lie on a circle

\text{C}

with

\text{PR}

as its diameter, The tangents to

\text{C}

at the points

\text{Q}

and

\text{R}

intersect at the point

\text{S}

. If

\text{S}

lies on the line

2\text{x} - \text{ky} = 1

, then

\text{k}

is equal to

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Step-by-step solutionView

Correct answer

The value of k is 3.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Circle $C$ has diameter $PR$ , where $P = (-3, 2)$ and $R = (a, 4)$ . Point $Q(9, 10)$ lies on $C$ . Tangents at $Q$ and $R$ intersect at $S$ . $S$ lies on $2x - ky = 1$ .

02goal

Find the value of $k$ .

03approach

Write circle $C$ in diametric form with endpoints $P(-3,2)$ and $R(a,4)$ . Substitute $Q(9,10)$ into the circle equation to find $a$ . Then write equations of tangents at $Q$ and $R$ using $T = 0$ , solve for intersection point $S(x_0, y_0)$ , and substitute $S$ into $2x - ky = 1$ to obtain $k$ .

04execute

Circle equation with diameter $PR$ : $(x + 3)(x - a) + (y - 2)(y - 4) = 0$ . Substitute $Q(9, 10)$ : $(9 + 3)(9 - a) + (10 - 2)(10 - 4) = 0 \implies 12(9 - a) + 8 \times 6 = 0 \implies 12(9 - a) + 48 = 0 \implies 9 - a = -4 \implies a = 13$ .

05execute

The circle is $(x + 3)(x - 13) + (y - 2)(y - 4) = 0$ , which expands to $x^2 - 10x - 39 + y^2 - 6y + 8 = 0$ , i.e., $x^2 + y^2 - 10x - 6y - 31 = 0$ .

06execute

Tangent at $Q(9, 10)$ : $9x + 10y - 5(x + 9) - 3(y + 10) - 31 = 0 \implies 4x + 7y - 45 - 30 - 31 = 0 \implies 4x + 7y = 106$ . Tangent at $R(13, 4)$ : $13x + 4y - 5(x + 13) - 3(y + 4) - 31 = 0 \implies 8x + y - 65 - 12 - 31 = 0 \implies 8x + y = 108$ .

07execute

Solve the system: from $8x + y = 108$ , $y = 108 - 8x$ . Substitute into $4x + 7y = 106$ : $4x + 7(108 - 8x) = 106 \implies 4x + 756 - 56x = 106 \implies -52x = -650 \implies x = 650/52 = 25/2$ . Then $y = 108 - 8(25/2) = 108 - 100 = 8$ . Thus, $S = (25/2, 8)$ .

08execute

Substitute $S(25/2, 8)$ into $2x - ky = 1$ : $2(25/2) - k(8) = 1 \implies 25 - 8k = 1 \implies 8k = 24 \implies k = 3$ .

✓verify

Check: for $k = 3$ , $2(25/2) - 3(8) = 25 - 24 = 1$ . The diameter condition is also verified since $\vec{PQ} = (12, 8)$ and $\vec{RQ} = (-4, 6)$ , giving $\vec{PQ} \cdot \vec{RQ} = 12(-4) + 8(6) = -48 + 48 = 0$ , confirming $\angle PQR = 90^\circ$ .

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does

(x+3)(x-a) + (y-2)(y-4) = 0

represent the circle?

Any point $(x, y)$ on a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ subtends a right angle at the diameter, which gives $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ via dot product of perpendicular vectors.

Why can we use the chord of contact formula here?

Since $S$ is the intersection of the tangents at $Q$ and $R$ , the line joining the points of contact $Q$ and $R$ is by definition the chord of contact of point $S$ .

Answer

The value of k is 3.

Why each option works or fails

Step-by-step solution

given: Circle $C$ has diameter $PR$ , where $P = (-3, 2)$ and $R = (a, 4)$ . Point $Q(9, 10)$ lies on $C$ . Tangents at $Q$ and $R$ intersect at $S$ . $S$ lies on $2x - ky = 1$ .
goal: Find the value of $k$ .
approach: Write circle $C$ in diametric form with endpoints $P(-3,2)$ and $R(a,4)$ . Substitute $Q(9,10)$ into the circle equation to find $a$ . Then write equations of tangents at $Q$ and $R$ using $T = 0$ , solve for intersection point $S(x_0, y_0)$ , and substitute $S$ into $2x - ky = 1$ to obtain $k$ .
execute: Circle equation with diameter $PR$ : $(x + 3)(x - a) + (y - 2)(y - 4) = 0$ . Substitute $Q(9, 10)$ : $(9 + 3)(9 - a) + (10 - 2)(10 - 4) = 0 \implies 12(9 - a) + 8 \times 6 = 0 \implies 12(9 - a) + 48 = 0 \implies 9 - a = -4 \implies a = 13$ .
execute: The circle is $(x + 3)(x - 13) + (y - 2)(y - 4) = 0$ , which expands to $x^2 - 10x - 39 + y^2 - 6y + 8 = 0$ , i.e., $x^2 + y^2 - 10x - 6y - 31 = 0$ .
execute: Tangent at $Q(9, 10)$ : $9x + 10y - 5(x + 9) - 3(y + 10) - 31 = 0 \implies 4x + 7y - 45 - 30 - 31 = 0 \implies 4x + 7y = 106$ . Tangent at $R(13, 4)$ : $13x + 4y - 5(x + 13) - 3(y + 4) - 31 = 0 \implies 8x + y - 65 - 12 - 31 = 0 \implies 8x + y = 108$ .
execute: Solve the system: from $8x + y = 108$ , $y = 108 - 8x$ . Substitute into $4x + 7y = 106$ : $4x + 7(108 - 8x) = 106 \implies 4x + 756 - 56x = 106 \implies -52x = -650 \implies x = 650/52 = 25/2$ . Then $y = 108 - 8(25/2) = 108 - 100 = 8$ . Thus, $S = (25/2, 8)$ .
execute: Substitute $S(25/2, 8)$ into $2x - ky = 1$ : $2(25/2) - k(8) = 1 \implies 25 - 8k = 1 \implies 8k = 24 \implies k = 3$ .
verify: Check: for $k = 3$ , $2(25/2) - 3(8) = 25 - 24 = 1$ . The diameter condition is also verified since $\vec{PQ} = (12, 8)$ and $\vec{RQ} = (-4, 6)$ , giving $\vec{PQ} \cdot \vec{RQ} = 12(-4) + 8(6) = -48 + 48 = 0$ , confirming $\angle PQR = 90^\circ$ .

Shortcut:

When to use it: Using $QR$ as the chord of contact of point $S$ with respect to circle $C$ .

given: Points $P(-3,2)$ , $Q(9,10)$ , and $R(a,4)$ lie on circle $C$ with diameter $PR$ . $S(x_1, y_1)$ is the intersection of tangents at $Q$ and $R$ .

goal: Find $k$ where $S$ lies on $2x - ky = 1$ .

approach: Since $PR$ is a diameter, $\angle PQR = 90^\circ$ . Use $\vec{PQ} \cdot \vec{RQ} = 0$ to get $a$ . The line $QR$ is the chord of contact from $S(x_1, y_1)$ to circle $C$ . Compare the equation of $QR$ with the line $T = 0$ from $S$ to circle $C$ to find $S$ , then find $k$ .

execute: $\vec{PQ} = (12, 8)$ and $\vec{RQ} = (9 - a, 6)$ . Since $\angle PQR = 90^\circ$ , $12(9 - a) + 8(6) = 0 \implies 9 - a = -4 \implies a = 13$ , so $R = (13, 4)$ .

execute: Line $QR$ passes through $(9, 10)$ and $(13, 4)$ : slope $m = (4 - 10)/(13 - 9) = -6/4 = -3/2$ . Equation of $QR$ : $y - 10 = -\frac{3}{2}(x - 9) \implies 3x + 2y = 47$ .

execute: The chord of contact of $S(x_1, y_1)$ with respect to $x^2 + y^2 - 10x - 6y - 31 = 0$ is $x x_1 + y y_1 - 5(x + x_1) - 3(y + y_1) - 31 = 0$ , which is $x(x_1 - 5) + y(y_1 - 3) = 5x_1 + 3y_1 + 31$ . Compare coefficients with $3x + 2y = 47$ : $\frac{x_1 - 5}{3} = \frac{y_1 - 3}{2} = \frac{5x_1 + 3y_1 + 31}{47}$ . Let this ratio be $\lambda$ . Then $x_1 = 3\lambda + 5$ , $y_1 = 2\lambda + 3$ . Substitute into the constant ratio: $5(3\lambda + 5) + 3(2\lambda + 3) + 31 = 47\lambda \implies 21\lambda + 65 = 47\lambda \implies 26\lambda = 65 \implies \lambda = 5/2$ . Thus $x_1 = 3(5/2) + 5 = 25/2$ and $y_1 = 2(5/2) + 3 = 8$ .

execute: Substitute $S(25/2, 8)$ into $2x - ky = 1$ : $2(25/2) - 8k = 1 \implies 25 - 8k = 1 \implies k = 3$ .

verify: Check: $2(12.5) - 3(8) = 25 - 24 = 1$ . The values match perfectly.

Circles: JEE Main Mathematics Question with Solution

What feels right?

The value of k is 3.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

08execute

✓verify

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