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 , If the line x – y + 2 = 0 intersects the circle at the points A and B, then is equal to: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(3)

x + (k – 1)y + 3 = 0

$\Rightarrow Slope, m_{1} = \frac{1}{1 - k}$

${(A B)}^{2}$

FIGUR

Now, $m_{1} m_{2} = - 1$

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

$\Rightarrow (1 - k) (k^{2}) - 2$

$\Rightarrow k^{3} - k^{2} + 2 = 0$

$\Rightarrow (k + 1) [k^{2} - 2 k + 2] = 0$

$\Rightarrow k = - 1$

$∴ x - 2 y + 3 = 0$

2x + y – 4 = 0

After sloving, we get x = 1, y = 2

$∴$ Centre is (1, 2)

Equation of circle is ${(x - 1)}^{2} + {(y - 2)}^{2} = r^{2}$

It passes through (0, 0)

$∴ {(0 - 1)}^{2} + {(0 - 2)}^{2} = r^{2}$

$\Rightarrow r^{2} = 5$

$∴ {(x - 1)}^{2} + {(y - 2)}^{2} = 5$

$\Rightarrow x^{2} + y^{2} - 2 x - 4 y = 0$

and equation of line x – y + 2 = 0

$p = \frac{1 - 2 + 2}{\sqrt{2}} = \frac{1}{\sqrt{2}}$

$A B = 2 \sqrt{r^{2} - p^{2}} = 2 \sqrt{5 - \frac{1}{2}} = 2 (3 / \sqrt{2}) = 3 / \sqrt{2}$

So, ${(A B)}^{2} = 18$

Solution

Q.
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: [2026]

1 10

2 27

3 18

4 34

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