Question

Let $P$ be point on the circle $x^2+y^2=9,Q$ a point on the line $7x+y-3=0$, and the perpendicular bisector of $PQ$ be the line $x-y+2=0$. Then coordinates of $P$ are

• A. $(0,-3)$
• B. $(0,3)$
• C. $(\frac{72}{25},-\frac{21}{25})$
• D. $(-\frac{72}{25},\frac{21}{25})$

Answer 1

The correct option is D $(-\frac{72}{25},\frac{21}{25})$

Let the coordinates of point $P$, which lies on circle $S:x^2+y^2=9$, be $(3\cos \theta ,3\sin \theta )$
Let the coordinates of point $Q$, which lies on line $L_1:7x+y=3$, be $(a,-7a-3)$
Let the mid-point of $PQ$ be $R(\frac{3\cos \theta +a}{2},\frac{3\sin \theta -7a-3}{2})$
Slope of line $PQ$ is $m_1=\frac{3\sin \theta +7a+3}{3\cos \theta -a}$
Perpendicular bisector of line $PQ$ is $L_2:x-y+2=0$
Slope of line $L_2$ is $m_2=1$
Point $R$ satisfies the line $L_2$, then
$\Rightarrow \frac{3\cos \theta +a}{2}-\frac{3\sin \theta -7a-3}{2}+1=0$
Which reduces to,
$3\cos \theta -3\sin \theta +8a+5=0$................... {i}
Since, $L_1$ and $PQ$ are perpendicular to each other
$\therefore m_1{m}_2=-1$
$m_1{m}_2=\frac{3\sin \theta +7a+3}{3\cos \theta -a}=-1$
Which gives,
$3\cos \theta +3\sin \theta +6a+3=0$................. {ii}
Solving {i} and {ii}}, we get
$3\sin \theta =a+1$ and
$3\cos \theta =-7a-4$
Squaring and adding these both, we get
$25a^2+29a+4=0$, which have $-1$ and $-\frac{4}{25}$ as roots.
If $a=-1$, then $\sin \theta =0$ and $\cos \theta =1$
$\therefore$ Points are $P(3,0)$ and $Q(-1,4)$
If $a=-\frac{4}{25}$, then $\sin \theta =\frac{7}{25}$ and $\cos \theta =-\frac{24}{25}$
$\therefore$ Points are $P(-\frac{72}{25},\frac{21}{25})$ and $Q(-\frac{4}{25},-\frac{47}{25})$
Hence, option D.