,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $$

Let the cordinate of centre of circle be $(g, f)$ which touches the line i.e. it will satisfy the equations of lines $L_1$ and $L_2$.

$$ \begin{array}{ll} \Rightarrow & g+f=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ \Rightarrow & g-f=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{array} $$

On solving Eqs. (iii) and (iv), we get (g. $f$ ) $=(2,0)$

The center of the required circle will lie on bisector of acute angle between these tangents i.e. on.

$$ \frac{x+y-2}{\sqrt{2}}=\frac{x-y-2}{\sqrt{2}} \text { i.e. } $$

Let it be ( $h, 0$ ) where $h$ is (positive). If $r$ be the radius, then it touches the circle $x^2+y^2=1$ externally

According to the condition,

$$ \begin{aligned} & C_1 C_2 =r_1+r_2\\& h =1+r \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(v)\\ \therefore \quad h & =2 \pm \sqrt{2 r}=1+r \end{aligned} $$

On solving, we get, $r=\sqrt{2}-1$

Hence, the equation of circle with center $(\sqrt{2}, 0)$ and radius $(\sqrt{2}-1)$ is

$$ \begin{aligned} & \Rightarrow(x-\sqrt{2})^2+y^2=(\sqrt{2}-1)^2 \\ & \Rightarrow(x-\sqrt{2})^2+y^2=3-2 \sqrt{2} \end{aligned} $$

Circle

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