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

Answer: (D)

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

Any point on $x^2+y^2=9$ is given by $P(3\cos \theta ,3\sin \theta )$, any point on $7x+y+3=0$ is given by $Q(a,-7a-3)$

Given that $L:x–y+1=0$ is perpendicular bisector of the line $PQ$

$⟹m_{PQ}\times m_L=-1$

$⟹m_{PQ}\times 1=-1$

$m_{PQ}=-1$

$\frac{-7a–3–3\sin \theta }{a–3\cos \theta }=-1$

$⟹3\cos \theta +3\sin \theta +6a+3=0$

$(\cos \theta +\sin \theta )=-2a-1–\left(1\right)$

And also midpoint of $PQ$ lies on $L$

$M=(\frac{3\cos \theta +a}{2},\frac{3\sin \theta –7a-3}{2})$

Substituting in $L$

$(3\cos \theta +a)–(3\sin \theta –7a–3)+2=0$

$⟹3\cos \theta –3\sin \theta +8a+5=0$

$\cos \theta -\sin \theta =\frac{-8a-5}{3}–\left(2\right)$

$(1{)}^2+(2{)}^2$

$1+2\sin \theta \cos \theta +1–2\sin \theta \cos \theta =(2a+1{)}^2+{\left(\frac{8a+5}{3}\right)}^2$

$⟹2\times 9–9(4a^2+4a+1)+(64a^2+80a+25)$

$⟹18=100a^2+34+116a$

$⟹25a^2+29a+4=0$

$(25a+4)(a+1)=0$

$a=-1,\frac{-4}{25}$

If $a=-1$ then from equation (1) and (2)

$\cos \theta +\sin \theta =1$

$\cos \theta -\sin \theta =1$

$⟹\theta =0,P=(3,0)$

If $a=\frac{-4}{25}$

$\cos \theta =\frac{-24}{25},\sin \theta =\frac{7}{25}⟹P=(\frac{-72}{25},\frac{21}{25})$