Question

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

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

Answer 1

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

Let the circle with tangent $2x+3y+1=0$ at $(1,-1)$ be

$(x-1{)}^2+(y+1{)}^2+\lambda (2x+3y+1)=0$

$x^2+y^2+(2\lambda -2)x+(3\lambda +2)+2+\lambda =0$

It is orthogonal to $x(x+2)+(y+1)(y-3)=0$

$x^2+y^2+2x-2y-3=0$

So, $2g_1{g}_2+2f_1{f}_2=c_1+c_2$

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

Equation of a required circle

$x^2+y^2+(2\left(\frac{-3}{2}\right)-2)x+[3\left(\frac{-3}{2}\right)+2]y+2-\frac{3}{2}=0$

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