Let $\mathrm{S}$ be the circumcircle of the triangle formed by the line $x-2 y-4=0$ with the coordinate axes. If $\mathrm{P}(-2,-4)$ is a point in the plane of the circle $S$ and $Q$ is a point on $S$ such that the distance between $\mathrm{P}$ and $\mathrm{Q}$ is the least, then $\mathrm{PQ}=$

Question

Aicharya · Accepted Answer

Line $x-2 y-4=0$ intersect the axes at $A(4,0)$ and $B(0,-2)$

$ \begin{aligned} & \therefore \text { Centre of circle }=\left(\frac{4+0}{2}, \frac{-2+0}{2}\right)=(2,-1) \\ & \text { radius }=\sqrt{(2-0)^2+(0+1)^2}=\sqrt{5} \\ & C P=\sqrt{(2+2)^2+(-1+4)^2}=5 \end{aligned} $ Distance of $P Q$ is least when points $P, Q$ and $C$ are collinear. $ \therefore \quad P Q=C P-C Q=5-\sqrt{5} $

Let $\mathrm{S}$ be the circumcircle of the triangle formed by the line $x-2 y-4=0$ with the coordinate axes…

Solution