The points of intersection of the line ax + by = 0 , ( a ≠ b ) and...

Given,

The points of intersection of the line $ax + by = 0, ( a \ne b )$ and the circle ${x}^{2} + {y}^{2} - 2x = 0$ are $A\left(\alpha, 0\right)$ and B ( 1, \beta )$then,</p><p>For point$A\left(\alpha, 0\right)$</p><p>$a ( \alpha ) + b ( 0 ) = 0 ⇒ \alpha = 0$</p><p>And,${\alpha}^{2} - 2\alpha = 0 ⇒ \alpha = 0, 2$</p><p>For point$B ( 1, \beta )$</p><p>$a + b\beta = 0$</p><p>Or,$\beta = \dfrac{- b}{a}$</p><p>And,$1 + {\beta}^{2} - 2 = 0$</p><p>$⇒ \beta = \pm 1$</p><p>$∵ a \ne b$therefore,</p><p>Only possibility$\alpha = 0, \beta = 1$</p><p>Points are$A\left(0, 0\right)$and$B\left(1, 1\right)$</p><p>So the centre of the circle with diameter AB is \left(\dfrac{1}{2}, \dfrac{1}{2}\right)$ and radius $r = \dfrac{1}{\sqrt{2}}$

Now the image of \left(\dfrac{1}{2}, \dfrac{1}{2}\right)$about the line$x + y + 2 = 0$is,</p><p>$\dfrac{x - \dfrac{1}{2}}{1} = \dfrac{y - \dfrac{1}{2}}{1} = \dfrac{- 2\left(\dfrac{1}{2} + \dfrac{1}{2} + 2\right)}{{1}^{2} + {1}^{2}}$</p><p>$⇒ \dfrac{x - \dfrac{1}{2}}{1} = \dfrac{y - \dfrac{1}{2}}{1} = - 3$</p><p>$⇒$Centre of image circle is \left(- \dfrac{5}{2}, - \dfrac{5}{2}\right)$

Equation of image circle

${\left(x + \dfrac{5}{2}\right)}^{2} + {\left(y + \dfrac{5}{2}\right)}^{2} = {\left(\dfrac{1}{\sqrt{2}}\right)}^{2}$

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