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(C) The lines $x+y=2$ and $x-y=2$ intersect at $(2, 0)$. Let the center of the required circle be $(h, k)$ and its radius be $r$. Since the circle tou

Vedclass · Accepted Answer

$(x-\sqrt{2})^2+y^2=(\sqrt{2}-1)^2$

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(C) The lines $x+y=2$ and $x-y=2$ intersect at $(2, 0)$. Let the center of the required circle be $(h, k)$ and its radius be $r$. Since the circle touches $x+y=2$ and $x-y=2$,the center must lie on the angle bisector of these lines,which are $x=2$ or $y=0$. Given the symmetry and the condition of touching $x^2+y^2=1$ (center $(0,0)$,radius $1$),the center lies on the $x$-axis,so $k=0$. The distance from $(h, 0)$ to $x+y-2=0$ is $r = \frac{|h+0-2|}{\sqrt{1^2+1^2}} = \frac{|h-2|}{\sqrt{2}}$. Since the circle touches $x^2+y^2=1$ externally or internally,the distance between centers is $d = |h-0| = |h|$. Condition for touching: $r = |h| \pm 1$. Case $1$: $r = h-1$. Then $\frac{h-2}{\sqrt{2}} = h-1 \implies h-2 = \sqrt{2}h - \sqrt{2} \implies h(\sqrt{2}-1) = \sqrt{2}-2$. $h = \frac{\sqrt{2}-2}{\sqrt{2}-1} = \frac{-\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1} = -\sqrt{2}$. Radius $r = |-\sqrt{2}-1| = \sqrt{2}+1$. Case $2$: $r = |h|+1$. $\frac{2-h}{\sqrt{2}} = h+1 \implies 2-h = \sqrt{2}h + \sqrt{2} \implies h(1+\sqrt{2}) = 2-\sqrt{2}$. $h = \frac{2-\sqrt{2}}{1+\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}+1} = \sqrt{2}(\sqrt{2}-1)^2 = \sqrt{2}(3-2\sqrt{2}) = 3\sqrt{2}-4$. Checking options,the circle with center $(\sqrt{2}, 0)$ and radius $\sqrt{2}-1$ satisfies the geometry. Thus,the equation is $(x-\sqrt{2})^2+y^2=(\sqrt{2}-1)^2$.

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

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