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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Aicharya · Accepted Answer

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 $m=\frac{1}{3}, 3 y=x$

If $y=3 x$ is a tangent to a circle with centre $(1,1)$, then the other tangent drawn through $(0,0)$ to the…

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