Mathematics - Circle Question with Solution | TestHub

Centre of circle $S_1=(2,4)$

Centre of circle $S_2=(4,2)$

Radius of circle $S_1=$ radius of circle $S_2=4$

∴ equation of circle ${\mathrm{S}}_2$

$(x-4{)}^2+(y-2{)}^2=16$

$\Rightarrow x^2+y^2-8x-4y+4=0$

Equation of circle touching $y=x$ at ( 1,1 ) can be taken as $(x-1{)}^2+(y-1{)}^2+\lambda (x-y)=0$

or, $x^2+y^2+x(\lambda -2)+y(-\lambda -2)+2=0$ ......(ii)

As this is orthogonal to ${\mathrm{S}}_2$

$\Rightarrow 2\left(\frac{\lambda -2}{2}\right)\cdot (-4)+2\left(\frac{-\lambda -2}{2}\right)\cdot (-2)=4+2$

$\Rightarrow -4\lambda +8+2\lambda +4=6$

∴ required equation of circle is

$x^2+y^2+x-5y+2=0$.

Radius $=\sqrt{\frac{1}{4}+\frac{25}{4}-2}=\sqrt{\frac{26-8}{4}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$.