Question

A variable chord of circle $x^2+y^2=4$ is drawn from the point $P(3,5)$ meeting the circle at the points $A$ and
$B$. A point $Q$ is taken on this chord such that $2PQ=PA+PB$. Locus of ${}^{\prime \prime }{Q}^{\prime \prime }$ is

• A. $x^2+y^2+3x+4y=0$
• B. $x^2+y^2=36$
• C. $x^2+y^2=16$
• D. $x^2+y^2-3x-5y=0$

Answer 1

The correct option is D $x^2+y^2-3x-5y=0$


Given, $2PQ=PA+PB$
$\Rightarrow PQ-PA=PB-PQ$
$\Rightarrow AQ=BQ$

So, $Q$ is midpoint of $AB$
Perpendicular from centre bisects the chord, so
$OQ\perp AB$
$\Rightarrow$ Slope of $OQ\times$ Slope of $AB=-1$

$\Rightarrow \frac{k-0}{h-0}\times \frac{k-5}{h-3}=-1\Rightarrow k^2-5k=-h^2+3h\Rightarrow k^2+h^2-5k-3h=0$

Hence, the locus of $Q$ is
$x^2+y^2-3x-5y=0$