Question

The equation of circle touching the line $2x+3y+1=0$ at $(1,-1)$ and cutting orthogonally the circle having line segment joining $(0,3)$ and $(-2,-1)$ as diameter 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 $2x^2+2y^2-10x-5y+1=0$

We are given that line
$2x+3y+1=0$ touches a circle $S=0$ $\text{at}(1,-1)$


So, equation of this circle can be given by $(x-1{)}^2+(y+1{)}^2+\lambda (2x+3y+1)=0,\lambda ϵR$
or $x^2+y^2+2x(\lambda -1)+y(3\lambda +2)+(\lambda +2)=\mathrm{0........}\left(i\right)$

But given that this circle is orthogonal to the circle, the extremities of whose diameter are $(0,3)$ and $(-2,-1)$

i.e., $x(x+2)+(y-3)(y+1)=0$
or $x^2+y^2+2x-2y-3=0$ ........$\left(ii\right)$

Applying the condition of orthogonality for Eqs. $\left(i\right)$ and $\left(ii\right)$, we get

$2(\lambda -1)\times 1+2\left(\frac{3\lambda +2}{2}\right)\times (-1)=\lambda +2-3$
$\Rightarrow 2\lambda -2-3\lambda -2=\lambda -1$
$\Rightarrow 2\lambda =-3$
$\Rightarrow \lambda =-\frac{3}{2}$

Substituting this value of $1$ in Eq. $\left(i\right)$ we get the required circle as
$x^2+y^2-5x-\frac{5}{2}y+\frac{1}{2}=0$

or $2x^2+2y^2-10x-5y+1=0$