Question

lf the equation of the circle passing through the origin and the points of intersection of the two circles $x^2+y^2-4x-6y-3=0$
$x^2+y^2+4x-2y-4=0$ is
$x^2+y^2+2ax+2by+c=0$ then the
ascending order of $a,b,c$ is

• A. $a,b,c$
• B. $b,c,a$
• C. $c,a,b$
• D. $a,c,b$

Answer 1

The correct option is A $a,b,c$

by system of circles,
equation of circle passing through point of intersection
of two circles $s_1$ and $s_2$ is given by
$s_1+\lambda s_2=0$
then $(x^2+y^2-4x-6y-3)+\lambda (x^2+y^2+4x-2y-4)=0$
Given, this circle pass through $(0,0)$
So, $-3+\lambda (-4)=0$
$\lambda =\frac{-3}{4}$
So, $e{q}^n$ of circle is
$\frac{x^2+y^2}{4}-7x-2y=0$
$x^2+y^2-28x-18y=0$
$a=-15,b=-9,$ and $c=0$
So, ascending order of $a,b,c$ is
$\Rightarrow a,b,c$