Question

Find the equation of the circle which touches the straight lines $x+y=2,x-y=2$ and also touches the circle $x^2+y^2=1$.

Answer 1

$l_2=0,l_2=0$ are tangent to required circle which intersect at $A(2,0)$. The centre of the required circle will lie on bisector of acute angle between these tangents i.e. on
$\frac{x+y-2}{\sqrt{2}}=\frac{x-y-2}{\sqrt{2}}$ i.e. $y=0$
Let it be $(h,0)$ where $h$ is +ive and if $r$ be the radius then it touches $x^2+y^2=1$ externally
$C_1{C}_2=r_1+r_2$ or $h=1+r.....\left(1\right)$
$p=r$ with any tangent gives $\frac{h-2}{\sqrt{2}}=r$
$\therefore h=2\pm \sqrt{2}r=1+r$, by $\left(1\right)$
$1=(1+\sqrt{2})r\therefore r=\frac{1}{1+\sqrt{2})}=\sqrt{2}-1$
Only as $r=+ive$
$\therefore h=2\pm \sqrt{2}(\sqrt{2}-1)=4-\sqrt{2},\sqrt{2}$
Hence the circles are $(4-\sqrt{2},0).(\sqrt{2}-1)$ and $(\sqrt{2},0),(\sqrt{2}-1)$.