Question

A circle touches the line $2x+3y+1=0$ at the point $(1,-1)$ and is orthogonal to the circle which has the line segment having end points $(0,-1)$ and $(-2,3)$ as diameter.Its equation is

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

Answer 1

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

The required circle is ${(x-1)}^2+{(y+1)}^2+\lambda (2x+3y+1)=0$
or $x^2+y^2+2(\lambda -1)x+(3\lambda +2)y+\lambda +2=0$ ...................$\left(1\right)$
The circle having the line joining $(0,-1)$ and $(-2,3)$ as diameter is
$x(x+2)+(y+1)(y-3)=0$
or $x^2+y^2+2x-2y-3=0$ ...................$\left(2\right)$
Eqns$\left(1\right)$ and $\left(2\right)$ are orthogonal.
$\therefore 2(\lambda -1)-(3\lambda +2)=\lambda +2-3$
On simplifying,
$\Rightarrow \lambda =-\frac{3}{2}$
Hence eqn$\left(1\right)$ becomes
$x^2+y^2+2(-\frac{3}{2}-1)x+(3\times -\frac{3}{2}+2)y-\frac{3}{2}+2=0$
$\Rightarrow x^2+y^2-5x-\frac{5}{2}y+\frac{1}{2}=0$