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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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The correct answer is:
$\sqrt{3}$
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 $(1,2)$
$=\sqrt{1^2+2^2-2}=\sqrt{3}$
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 $(1,2)$
$=\sqrt{1^2+2^2-2}=\sqrt{3}$
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