Question

Let P be point on the circle $x^2+y^2=9$ , Q a point on 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 $(-\frac{72}{25},\frac{21}{25})$

$x^2+y^2=9$ $P(3cos\theta ,3sin\theta )$

$7x+y+3=0$ $(t,-3-3t)$

$\left[\right(3cos\theta +t){]}_2,(3sin\theta -3-7t\left)/2\right)$ $x-y+1=0$

$\frac{3cos\theta }{2}+\frac{t}{2}-\frac{-3sin\theta }{2}+\frac{3}{2}+\frac{7t}{2}+1=0$

$m_1{m}_2=-1$ $PQ=\frac{3sin\theta +3+7t}{3cos\theta -t}=m_1$

$m_1\times 1=-1$ $\frac{3sin\theta +3+7t}{3cos\theta -t}=-1$

$3sin\theta +3+7t=t-3cos\theta$

$3(sin\theta +cos\theta )+8t+3=\mathrm{0...}\left(2\right)$

$3(sin\theta +cos\theta )+8t+5=\mathrm{0...}\left(1\right)$ square & add

$3^2\left(1\right)+3^2\left(1\right)=100t^2+116t+34$

$100t^2+116t+26=0$ $t=-1,t=-4/25$

$3(cos\theta -sin\theta )=\frac{32}{25}-5=-93/25$

$(\frac{-72}{25},\frac{21}{25})$