The points of intersection of the line ax+by=0, (a≠ b) and the circle x^2+y^2-2x=0 are A(α ,0) and B(1,β ). The image of the circle with AB as a…

Given,

The points of intersection of the line $ax+by=0,(a\neq b)$ax+by=0,(a=b) and the circle ${x}^{2}+{y}^{2}-2x=0$x2+y2−2x=0 are $A(\alpha ,0)$A(α,0) and $B(1,\beta )$B(1,β) then,

For point $A(\alpha ,0)$A(α,0)

$$a(\alpha )+b(0)=0\Rightarrow \alpha =0$$a(α)+b(0)=0⇒α=0

And, ${\alpha }^{2}-2\alpha =0\Rightarrow \alpha =0,2$α2−2α=0⇒α=0,2

For point $B(1,\beta )$B(1,β)

$$a+b\beta =0$$a+bβ=0

Or, $\beta =\frac{-b}{a}$β=a−b​

And, $1+{\beta }^{2}-2=0$1+β2−2=0

$$\Rightarrow \beta =\pm 1$$⇒β=±1

$\because a\neq b$∵a=b therefore,

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

Points are $A(0,0)$A(0,0) and $B(1,1)$B(1,1)

So the centre of the circle with diameter AB is $(\frac{1}{2},\frac{1}{2})$(21​,21​) and radius $r=\frac{1}{\sqrt{2}}$r=2​1​

Now the image of $(\frac{1}{2},\frac{1}{2})$(21​,21​) about the line $x+y+2=0$x+y+2=0 is,

$$\frac{x-\frac{1}{2}}{1}=\frac{y-\frac{1}{2}}{1}=\frac{-2(\frac{1}{2}+\frac{1}{2}+2)}{{1}^{2}+{1}^{2}}$$1x−21​​=1y−21​​=12+12−2(21​+21​+2)​

$$\Rightarrow \frac{x-\frac{1}{2}}{1}=\frac{y-\frac{1}{2}}{1}=-3$$⇒1x−21​​=1y−21​​=−3

$\Rightarrow$⇒Centre of image circle is $(-\frac{5}{2},-\frac{5}{2})$(−25​,−25​)

Equation of image circle

$${(x+\frac{5}{2})}^{2}+{(y+\frac{5}{2})}^{2}={(\frac{1}{\sqrt{2}})}^{2}$$(x+25​)2+(y+25​)2=(2​1​)2

$$\Rightarrow {x}^{2}+{y}^{2}+5x+5y+12=0$$⇒x2+y2+5x+5y+12=0