Question

The equation of circle touching the line $2x+3y+1=0$ at $(1,-1)$ and orthogonally cutting the cirlce whose endpoints of diameter are $(0,3)$ and $(-2,-1)$ is

• A. $2x^2+2y^2+10x+5y+1=0$
• B. $2x^2+2y^2-10x-5y+1=0$
• C. $2x^2+2y^2-10x-10y+10=0$
• D. $2x^2+2y^2-10x-10y+20=0$

Answer 1

The correct option is B $2x^2+2y^2-10x-5y+1=0$

Given line $2x+3y+1=0$


So, the equation of circle $S$ is
$(x-1{)}^2+(y+1{)}^2+\lambda (2x+3y+1)=0,\lambda \in R\Rightarrow x^2+y^2+x(2\lambda -2)+y(3\lambda +2)+(\lambda +2)=0\cdots \left(i\right)$

Equation of circle whose endpoints of diameter are $(0,3)$ and $(-2,-1)$ is
$x(x+2)+(y-3)(y+1)=0\Rightarrow x^2+y^2+2x-2y-3=0\cdots \left(ii\right)$

Circle $\left(i\right)$ and $\left(ii\right)$ are orthogonal, so
$2(\lambda -1)\times 1+2\left(\frac{3\lambda +2}{2}\right)\times (-1)=\lambda +2-3\Rightarrow 2\lambda -2-3\lambda -2=\lambda -1\Rightarrow \lambda =-\frac{3}{2}$

Substituting $\lambda$ in equation $\left(i\right)$, we get
$x^2+y^2-5x-\frac{5}{2}y+\frac{1}{2}=0$

Hence, the equation of circle is
$2x^2+2y^2-10x-5y+1=0$