The points of intersection of the line a x + b y = 0 ( a b ) and the circle x 2 + y 2 - 2 x = 0 are A (, 0 ) and B ( 1 ,) . The image of the circle with A B as a diameter in the line x + y + 2 = 0 is [2023]

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Solution

Q.
The points of intersection of the line $a x + b y = 0 (a \neq b)$ and the circle $x^{2} + y^{2} - 2 x = 0$ are $A (α, 0)$ and $B (1, β)$ . The image of the circle with $A B$ as a diameter in the line $x + y + 2 = 0$ is [2023]

1 $x^{2} + y^{2} + 3 x + 3 y + 4 = 0$

2 $x^{2} + y^{2} + 5 x + 5 y + 12 = 0$

3 $x^{2} + y^{2} + 3 x + 5 y + 8 = 0$

4 $x^{2} + y^{2} - 5 x - 5 y + 12 = 0$

Ans.
(2)

Clearly the line and circle intersect at (0, 0)

$\Rightarrow α = 0$

$(1, β)$ lies on the circle

$1 + β^{2} - 2 \times 1 = 0 \Rightarrow β^{2} = 1 \Rightarrow β = \pm 1$

Let $β = - 1$

(1, -1) should lie on the line

$\Rightarrow a - b = 0 \Rightarrow a = b but a \neq b$

Therefore $β = 1$ .

The line and circle intersect at (0, 0) and (1, 1).

Centre of circle with diameter AB is $(\frac{1}{2}, \frac{1}{2})$

$Radius of circle = \frac{1}{\sqrt{2}}, where A (0, 0) and B (1, 1)$

Let (p, q) be the mirror image of $(\frac{1}{2}, \frac{1}{2})$ w.r.t. $x + y + 2 = 0$

$\frac{p - 1 / 2}{1} = \frac{q - 1 / 2}{1} = \frac{- 2 (\frac{1}{2} + \frac{1}{2} + 2)}{2} = - 3$ $\Rightarrow p = \frac{- 5}{2}, q = \frac{- 5}{2}$

(p, q) is centre of required circle whose equation is

${(x + \frac{5}{2})}^{2} + {(y + \frac{5}{2})}^{2} = \frac{1}{2} \Rightarrow x^{2} + y^{2} + 5 x + 5 y + \frac{25}{2} - \frac{1}{2} = 0$

$\Rightarrow x^{2} + y^{2} + 5 x + 5 y + 12 = 0$

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