The points of intersection of the line a x   +   b y = 0 , ( a ≠ b ) and the circle x 2 + y…

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,

For point $A\left(\alpha ,0\right)$

$a\left(\alpha \right)+b\left(0\right)=0\Rightarrow \alpha =0$

And, ${\alpha }^2-2\alpha =0\Rightarrow \alpha =0,2$

For point $B(1,\beta )$

$a+b\beta =0$

Or, $\beta =\frac{-b}{a}$

And, $1+{\beta }^2-2=0$

$\Rightarrow \beta =\pm 1$

$\because a\ne b$ therefore,

Only possibility $\alpha =0, \beta =1$

Points are $A\left(0,0\right)$ and $B\left(1,1\right)$

So the centre of the circle with diameter AB is $\left(\frac{1}{2},\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)$ and radius $r=\frac{1}{\sqrt{2}}$

Now the image of $\left(\frac{1}{2},\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)$ about the line $x+y+2=0$ is,

$\frac{x-{\displaystyle \frac{1}{2}}}{1}=\frac{{\displaystyle y-\frac{1}{2}}}{1}=\frac{-2\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}+2\right)}{1^2+1^2}$

$\Rightarrow \frac{x-{\displaystyle \frac{1}{2}}}{1}=\frac{{\displaystyle y-\frac{1}{2}}}{1}=-3$

$\Rightarrow$Centre of image circle is $\left(-\frac{5}{2},-\frac{5}{2}\right)$

Equation of image circle

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

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