Question

If $P$ is a point on the circle $x^2+y^2=9,Q$ is a point on the line $7x+y+3=0$, and the line $x-y+1=0$ is the perpendicular bisector of $PQ$, then the coordinates of $P$ are

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

Answer 1

Answer: (A), (D)

The correct options are
A $(3,0)$
D $(-\frac{72}{25},\frac{21}{25})$

Let coordinates of $P=(3cos\alpha ,3sin\alpha )$
Let $x-$ coordinate of Q is $x_1$ then
$y-$ coordinate of $Q$ is $-7x_1-3$ then
$\therefore Q=(x_1,-7x_1-3)$
$\because x-y+1=0$ is the perpendicular bisector of $PQ$ then midpoint of $PQ$ lie on $x-y+1=0$
$\Rightarrow \frac{3\cos \alpha +x_1}{2}-\frac{3\sin \alpha -7x_1-3}{2}+1=0$
$\Rightarrow 8x_1+3\cos \alpha -3\sin \alpha +5=0$
Multiply the above equation by $3$
$\Rightarrow 24x_1+9\cos \alpha -9\sin \alpha +15=0$ ..........$\left(1\right)$
and slope of$(x-y+1=0)\times$slope of $PQ=-1$
$\Rightarrow 1\times \frac{3\sin \alpha +7x_1+3}{3\cos \alpha -x_1}=-1$
$\Rightarrow 3\sin \alpha +7x_1+3=-3\cos \alpha +x_1$
$\Rightarrow 6x_1+3\sin \alpha +3\cos \alpha +3=0$
Multiply the above equation by $4$
$\Rightarrow 24x_1+12\sin \alpha +12\cos \alpha +12=0$ ..............$\left(2\right)$
Subtracting $\left(1\right)$ and $\left(2\right)$, we obtain
$-3\cos \alpha -21\sin \alpha +3=0$
$\Rightarrow 3(1-\cos \alpha )=21\sin \alpha$
$\Rightarrow (1-\cos \alpha )=7\sin \alpha$
Squaring both sides, we get
$\Rightarrow {(1-\cos \alpha )}^2=49{\sin }^2\alpha$
$\Rightarrow {(1-\cos \alpha )}^2=49(1-{\cos }^2\alpha )$
$\Rightarrow {(1-\cos \alpha )}^2=49(1+\cos \alpha )(1+\cos \alpha )$
Removing the common both sides, we get
$\Rightarrow (1-\cos \alpha )(1-\cos \alpha -49-49\cos \alpha )$
$\therefore \cos \alpha =1$ and $-50\cos \alpha =48$
or $\cos \alpha =1$ and $\cos \alpha =-\frac{24}{25}$
$\therefore$ Co-ordinates of $P$ are $(3,0)$ and $(-\frac{72}{25},\frac{21}{25})$