Question

If $P$ and $Q$ are two points on the circle $x^2+y^2-4x-4y-1=0$ which are farthest and nearest respectively from the point $(6,5)$ then

• A. $P=(-\frac{22}{5},3)$
• B. $Q=(\frac{22}{5},\frac{19}{5})$
• C. $P=(\frac{14}{3},-\frac{11}{5})$
• D. $Q=(-\frac{14}{3},-4)$

Answer 1

The correct option is B $Q=(\frac{22}{5},\frac{19}{5})$

We have $x^2+y^2-4x-4y-1=0$
Centre $C(2,2)$ radius $=\sqrt{(4+4+1)}=3$
Let $A(6,5)$
$CA=\sqrt{\{{(6-2)}^2+{(5-2)}^2\}}=5$

$P$ is on the circle, then $CP=3$
$\Rightarrow AP=CA-CP=5-3=2$
$CP:PA=3:2$
$x=\frac{3\times 6+2\times 2}{3+2}=\frac{18+4}{5}=\frac{22}{5}$, $y=\frac{3\times 5+2\times 2}{3+2}=\frac{15+4}{5}=\frac{19}{5}$
$\therefore P(\frac{22}{5},\frac{19}{5})$
For $Q,AQ:AC$
$AQ=AC+CQ=5+3=8$
$\therefore AQ:AC=8:5$
$x=\frac{8\times 2-5\times 6}{8-5}=\frac{16-30}{3}=-\frac{14}{3}$, $y=\frac{8\times 2-5\times 5}{8-5}=\frac{16-25}{3}=-\frac{9}{3}=-3$
$\therefore Q(-\frac{14}{3},-3)$