Question

lf $(2,3)$ and $(-3,2)$ are the limiting points of a coaxial system whose equation is
$(x^2+y^2-4x-6y+13)+\lambda (ax+by+c)=0$ then $a+b=$

• A. $6$
• B. $-6$
• C. $4$
• D. $-4$

Answer 1

The correct option is A $6$

$(x^2+y^2-(4x-6y+13\left)\right)+\lambda (ax+3y+c)=0$ is a coaxial system
of circle whose radical axis is
$ax+by+c=0$
and $ax+by+c=0$ is perpendicular to line joining $(2,3)$
and $(-3,2)$
so, $\frac{-1}{\frac{1}{5}}=\frac{-a}{b}$
$\Rightarrow -5=\frac{-a}{b}$
$\Rightarrow \frac{a}{b}=5$
So, let $a=5$ and $b=1$
$\Rightarrow a+b=6$