Question

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+3y+2=0$, $x^2+y^2+2x-3y-4=0$ is

• A. $x^2+y^2+2x+2y+2=0$
• B. $x^2+y^2+2x+2y-1=0$
• C. $x^2+y^2+2x+2y+1=0$
• D. $x^2+y^2+2x+2y+3=0$

Answer 1

The correct option is C $x^2+y^2+2x+2y+1=0$

Common chord of given circles is $S_1-S_2=0$

$\therefore 6y+6=0$

$y=-1$ is comman chord.

Now, we have to calculate a length of common chord of given circles.

Equation of $AB$ is $y+1=0$

$\therefore AE=\left|\frac{\frac{3}{2}+1}{1}\right|=\frac{5}{2}$

$\therefore AE=\sqrt{\frac{29}{4}-\frac{25}{4}}=1$

So, length of chord $2$.

So, the equation of the circle is

$\Rightarrow (x+1{)}^2+(y+1{)}^2=1$

$\Rightarrow x^2+y^2+2x+2y+1=0$

Foot of perpendicular from $(-1,\frac{3}{2})toy+1=0$

$\frac{h+1}{0}=\frac{k+\frac{3}{2}}{1}=-1\frac{(3+1)}{1}$

$h=-1,k=-1$