Family of circles with $AB$ as diameter

Question

The circle $S_1 :x^2+y^2-4=0$ cuts the circle $S_2 :x^2+y^2+2x+3y-5=0$ in A and B. Then find the equation of circle with $AB$ as diameter.

Answer is $13(x^2+y^2)-4x-6y-50=0$

Equation of AB will $S_2-S_1$ and then I used the fact that centre of one of the circles in the family of circles $S_1+kS_2=0$ must lie on $AB$ but I am not getting the correct answer. Any suggestions?

Answer 1

Your reasoning is correct, so I suspect you just made an algebra error.

As you say, the line $AB$ has equation $S_1 - S_2=0$, which is $$ 2x +3y - 1 = 0 $$ Any circle $C_k$ through $A$ and $B$ has equation $S_1 + kS_2=0$ for some $k$. Doing the algebra tells us that $C_k$ has equation $$ (1+k)x^2 + (1+k)y^2 +2x +3y -5 - 4k = 0 $$ The center of $C_k$ is at the point $$ P_k = - \left( \frac{2}{2(1+k)}, \frac{3}{2(1+k)} \right) $$ If $C_k$ has diameter $AB$, then $P_k$ lies on $AB$, so $$ -\frac{4}{2(1+k)} - \frac{9}{2(1+k)} = 1 $$ This gives $1+k = -13/2$ and $k = -15/2$. Substituting back in the equation of $C_k$ gives $$ 13(x^2+y^2)-4x-6y-50=0 $$

Answer 2

Write as $S_1 :x^2+y^2=4$ and $S_2 :x^2+y^2=-2x-3y+5$

Now to find the equation of AB: $S_1=S_2$

$4=-2x-3y+5\Longrightarrow \text{ the equation of line AB: }y=\frac 1 3 -\frac{2x}{3}\\ \text{let's find A,B: } (\frac 1 3 -\frac{2x}{3})^2+x^2=4\Longrightarrow x_{1,2}=\frac{2}{13}\pm\frac{3\sqrt{51}}{13}\\ (\frac{2}{13}+\frac{3\sqrt{51}}{13})^2+y^2=4\Longrightarrow y_{1,2}=\pm\frac{1}{13}\sqrt{3(71-4\sqrt{51})}\\ \text{now find the center $a,b$ using } a=\frac{x_1+x_2}{2}\text{ and } b=\frac{y_1+y_2}{2}\\ (x-a)^2+(y-b)^2=R^2\\ $