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If $y=3 x$ is a tangent to a circle with centre $(1,1)$, then the other tangent drawn through $(0,0)$ to the circle is
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Verified Answer
The correct answer is:
$3 y=x$
Since the line $y-3 x=0$ touches the circle $\therefore$ radius $=$ perpendicular distance from the centre $(1,1)$ to the tangent
Let the other equation of tangent which is passing through origin is $y=m x$.
$\text { radius }=\frac{|1-m|}{\sqrt{1+m^2}}$
$\begin{aligned}
\Rightarrow & \frac{4}{10}=\frac{(1-m)^2}{\left(1+m^2\right)} \\
\Rightarrow & 2\left(1+m^2\right)=5\left(1+m^2-2 m\right) \\
\Rightarrow & 3 m^2-10 m+3=0 \\
\Rightarrow & 3 m^2-9 m-m+3=0 \\
\Rightarrow & (3 m-1)(m-3)=0 \\
\Rightarrow & m=3, \frac{1}{3}
\end{aligned}$
at $m=3, y=3 x$ it is already given. at $m=\frac{1}{3}, 3 y=x$
Let the other equation of tangent which is passing through origin is $y=m x$.
$\text { radius }=\frac{|1-m|}{\sqrt{1+m^2}}$
$\begin{aligned}
\Rightarrow & \frac{4}{10}=\frac{(1-m)^2}{\left(1+m^2\right)} \\
\Rightarrow & 2\left(1+m^2\right)=5\left(1+m^2-2 m\right) \\
\Rightarrow & 3 m^2-10 m+3=0 \\
\Rightarrow & 3 m^2-9 m-m+3=0 \\
\Rightarrow & (3 m-1)(m-3)=0 \\
\Rightarrow & m=3, \frac{1}{3}
\end{aligned}$
at $m=3, y=3 x$ it is already given. at $m=\frac{1}{3}, 3 y=x$
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