Maths Q10 – JEE-MAIN 2026

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines

x + (k - 1) y + 3 = 0

and

2 x + k^{2} y - 4 = 0

. If the line

x - y + 2 = 0

intersects the circle at the points A and B, then

(A B)^{2}

is equal to :

Clue 1 of 3: Perpendicular Lines and Circle Properties

The problem involves finding the center and radius of a circle, and then the length of a chord. Recall the condition for two lines to be perpendicular and the distance formula.

🧠 Full Solution Path

Step 1: Determine the value of k✦ Active

The slopes of the lines $x + (k - 1) y + 3 = 0$ and $2 x + k^{2} y - 4 = 0$ are $m_{1} = - \frac{1}{k - 1}$ and $m_{2} = - \frac{2}{k^{2}}$ respectively. Since the lines are mutually perpendicular, their product of slopes is $- 1$ .

(- \frac{1}{k - 1}) (- \frac{2}{k^{2}}) = - 1

\frac{2}{k^{2} (k - 1)} = - 1 ⟹ 2 = - k^{3} + k^{2} ⟹ k^{3} - k^{2} + 2 = 0

By inspection, $k = - 1$ is a root: $(- 1)^{3} - (- 1)^{2} + 2 = - 1 - 1 + 2 = 0$ . Factoring the cubic equation gives $(k + 1) (k^{2} - 2 k + 2) = 0$ . The quadratic factor $k^{2} - 2 k + 2$ has a discriminant $D = (- 2)^{2} - 4 (1) (2) = - 4 < 0$ , so it has no real roots. Thus, $k = - 1$ is the only real solution.

Step 2: Find the center and radius of the circle○ Expand

Substitute $k = - 1$ into the line equations to find the center $(h, k)$ of the circle:

Line 1: x - 2 y + 3 = 0

Line 2: 2 x + y - 4 = 0

Solving this system of equations (e.g., multiply Line 2 by 2 and add to Line 1) yields $x = 1, y = 2$ . So, the center of the circle is $C (1, 2)$ . Since the circle passes through the origin $(0, 0)$ , the radius squared $R^{2}$ is the square of the distance from $C (1, 2)$ to $(0, 0)$ :

R^{2} = (1 - 0)^{2} + (2 - 0)^{2} = 1^{2} + 2^{2} = 1 + 4 = 5

Step 3: Calculate the square of the chord length

(A B)^{2}

○ Expand

The line $x - y + 2 = 0$ intersects the circle at points A and B. The perpendicular distance $d$ from the center $C (1, 2)$ to this line is given by the formula:

d = \frac{| 1 - 2 + 2 |}{\sqrt{1^{2} + (- 1)^{2}}} = \frac{| 1 |}{\sqrt{2}} = \frac{1}{\sqrt{2}}

So, $d^{2} = {(\frac{1}{\sqrt{2}})}^{2} = \frac{1}{2}$ . The length of the chord $A B$ is $2 \sqrt{R^{2} - d^{2}}$ . Therefore, $(A B)^{2} = 4 (R^{2} - d^{2})$ . Substituting the values of $R^{2}$ and $d^{2}$ :

(A B)^{2} = 4 (5 - \frac{1}{2}) = 4 (\frac{10 - 1}{2}) = 4 (\frac{9}{2}) = 2 \times 9 = 18

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Maths Question 10 – JEE-MAIN 2026

Clue 1 of 3: Perpendicular Lines and Circle Properties

🧠 Full Solution Path