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 $x-y+1=0$ . Then the coordinate 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 $P(x_1,y_1)$ and $Q(x_2,y_2)$

be two points on the circle $x^2+y^2=9$ and the line $7x+y+3=0$ respectively

$\therefore y_2=-(7x_2+3)$

Since, $x-y+1=0$ is perpendicular bisector of $PQ$

$\therefore \frac{1}{2}(x_1+x_2)-\frac{1}{2}[y_1-(7x_2+3)]+1=0$

$\Rightarrow x_1-y_1=-({8x}_2+5)$ ...(1)

And $m_1\times m_2=-1$

$\Rightarrow 1.\frac{(y_1+{7x}_2+3)}{(x_1-x_2)}=-1$

$\Rightarrow x_1+y_1=-({6x}_2+3)$ ...(2)

Squaring $\left(1\right)$and $\left(2\right)$and adding

$2(x_1^2+y_1^2)=100x_2^2+116x_2+34$

$\Rightarrow 100x_2^2+116x_2+16=0$

$\Rightarrow x_2=-1,\frac{-4}{25}$ ...(3)

Hence from (1) and (2) we get coordinates of $P$ as $P(3,0)$ and $(\frac{-72}{25},\frac{21}{25})$