Question

The equation of the circle passing through the point of intersection of the circle $x^2+y^2+2x+3y+1=0,$ $x^2+y^2+4x+3y+2=0$ and through the point $(-1,2)$ is

• A. $9x^2+92+16y-34=0$
• B. $10(x^2+y^2)+3y-86=0$
• C. $x^2+y^2-x-2y=0$
• D. $x^2+y^2+22x+12y+11=0$

Answer 1

Answer: (D)

Equation of circle through the intersection of the circle $S_1\equiv x^2+y^2+2x+3y+1=0$

$S_2\equiv x^2+y^2+4x+3y+2=0$

$S_1+\lambda S_2=0\Rightarrow x^2+y^2+2x+3y+1+\lambda (x^2+y^2+4x+3y+2)=0$

And it passes through $(-1,2)$.

$\therefore x^2+y^2+2x+3y+1+\lambda (x^2+y^2+4x+3y+2)\Rightarrow {(-1)}^2+{\left(2\right)}^2+2(-1)+3\left(2\right)+1+\lambda ({(-1)}^2+{\left(2\right)}^2+4(-1)+3(2)+2)=0\Rightarrow 1+4-2+6+1+\lambda (1+4-4+6+2)=0\Rightarrow 10+\lambda \left(9\right)=0\Rightarrow \lambda =\frac{-10}{9}$

$\therefore$ Equation of circle is

$x^2+y^2+2x+3y+1-\frac{10}{9}(x^2+y^2+4x+3y+2)\Rightarrow 9x^2+9y^2+18x+18y+9-10x^2-10y^2-40x-30y-20=0\Rightarrow x^2+y^2+22x+12y+11=0$