Question

If $(1,-2)$ is a pole of the circle $x^2+y^2-10x-10y+25=0$, then the equation of the diameter which bisects the polar is

• A. $2x+y-15=0$
• B. $4x-7y+15=0$
• C. $x-2y+5=0$
• D. $7x-4y-15=0$

Answer 1

The correct option is D $7x-4y-15=0$

Given circle is
$x^2+y^2-10x-10y+25=0$
$C=(5,5),r=5$
Equation of polar when pole is $(1,-2)$ is
$x-2y-5(x+1)-5(y-2)+25=0\Rightarrow -4x-7y+30=0\Rightarrow 4x+7y-30=0$
Distance from centre to polar is
$\frac{|20+35-30|}{\sqrt{65}}=\frac{25}{\sqrt{65}}<5$
So, polar is a chord
The diameter which bisects the chord is perpendicular to it, so
$7x-4y+\lambda =0$
It passes through centre $(5,5)$
$\Rightarrow 35-20+\lambda =0\Rightarrow \lambda =-15$

Hence, the required equation of the diameter is
$7x-4y-15=0$