A circle $S=0$ with radius $\sqrt{2}$ touches the line $x+y-2=0$ at $(1,1)$. Then, the length of the tangent drawn from the point $(1,2)$ to $S=0$ is

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Aicharya · Accepted Answer

Equation of line at $(1,1)$ is $x+y-2=0$ Slope of this line is -1 . So, slope of line perpendicular to this line is 1 . $\therefore \quad \tan \theta=1 \quad \theta=\frac{\pi}{4}$

Let, centre of circle $(h, k)$ i.e. $x=h \pm r \cos \theta$ and $y=k \pm r \sin \theta$ As, it passes through $(1,1)$. $\begin{aligned} & \therefore \quad h=1 \pm \sqrt{2} \cos \frac{\pi}{4} \\ & k=1 \pm \sqrt{2} \sin \frac{\pi}{4} \\ & \Rightarrow \quad h=1 \pm \frac{\sqrt{2}}{\sqrt{2}} \text { and } k=1 \pm \frac{\sqrt{2}}{\sqrt{2}} \\ & \Rightarrow \quad h=2,0 \text { and } k=2,0 \end{aligned}$ $\therefore \quad$ Centres are $(2,2)$ or $(0,0)$ Hence, equation of circle $x^2+y^2=2$ or $(x-2)^2+(y-2)^2=2$ $\therefore \quad$ Length of tangent from $(1,2)$ $=\sqrt{1^2+2^2-2}=\sqrt{3}$

A circle $S=0$ with radius $\sqrt{2}$ touches the line $x+y-2=0$ at $(1,1)$. Then, the length of the tangent…

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