JEE Main 2017 (Online) 8th AprilQuestion 1
If a point $P(0,-2)$ and $Q$ is any point on the circle $x^{2}+y^{2}-5x-y+5=0$, then the maximum value of $(PQ)^{2}$ is :
Circle
Solution
Concept. The farthest point $Q$ of a circle from $P$ is on line $PC$, beyond $C$: $PQ_{\max}=PC+r$. Formulas used. (1) Centre $(-g,-f)$, $r=\sqrt{g^2+f^2-c}$.(2) $PQ_{\max}=PC+r$.Step 1 - Circle, use (1). Centre $C=\left(\frac52,\frac12\right)$, $r^2=\frac{25}4+\frac14-5=\frac32$, $r=\sqrt{\frac32}$. Step 2 - $PC$. $P=(0,-2)$: $PC^2=\frac{25}4+\frac{25}4=\frac{25}2$, $PC=\frac5{\sqrt2}$. Step 3 - Use (2) and square. $$PQ_{\max}^2=\left(\frac5{\sqrt2}+\frac{\sqrt3}{\sqrt2}\right)^2=\frac{25+3+10\sqrt3}{2}=14+5\sqrt3.$$ Answer: C