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(C) Let $r$ be the radius of the required circle.From the geometry,the center of the circle lies on the $x$-axis at $(h, 0)$.The lines $x+y=2$ and $x-

Vedclass · Accepted Answer

$(x-\sqrt{2})^2+y^2=3-2\sqrt{2}$

Vedclass · Answer

(C) Let $r$ be the radius of the required circle.From the geometry,the center of the circle lies on the $x$-axis at $(h, 0)$.The lines $x+y=2$ and $x-y=2$ intersect at $P(2, 0)$.The distance from the center $(h, 0)$ to the line $x+y-2=0$ is equal to the radius $r$.$\frac{|h+0-2|}{\sqrt{1^2+1^2}} = r \Rightarrow \frac{|h-2|}{\sqrt{2}} = r$.Since the circle is to the left of $P(2, 0)$,$h Also,the circle touches $x^2+y^2=1$ externally,so the distance between centers is $r_1+r_2$.Distance between $(h, 0)$ and $(0, 0)$ is $h = 1+r$.Equating the two expressions for $h$:$1+r = 2 - r\sqrt{2}$$r(1+\sqrt{2}) = 1$$r = \frac{1}{\sqrt{2}+1} = \sqrt{2}-1$.Then $h = 1 + (\sqrt{2}-1) = \sqrt{2}$.The equation of the circle is $(x-\sqrt{2})^2 + y^2 = r^2 = (\sqrt{2}-1)^2 = 2+1-2\sqrt{2} = 3-2\sqrt{2}$.

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

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