The equation of the circle whose end points of a diameter are the centres of the circles $x^{2}+y^{2}+2 x-4…
The equation of the circle whose end points of a diameter are the centres of the
circles $x^{2}+y^{2}+2 x-4 y+1=0$ and $x^{2}+y^{2}-8 x+6 y+17=0$ is
$x^{2}+y^{2}-3 x-y-10=0$
$x^{2}+y^{2}+3 x-y-10=0$
$x^{2}+y^{2}+3 x+y-10=0$
$x^{2}+y^{2}-3 x+y-10=0$
Solution
Let centres of given circles be $A(-1,2)$ and $B(4,-3)$
By diameter form of equation of circle, we write
$\begin{array}{l}(x+1)(x-4)+(y-2)(y+3)=0 \\ \therefore x^{2}-4 x+x-4+y^{2}+3 y-2 y-6 \\ \therefore x^{2}+y^{2}-3 x+y-10=0\end{array}=0$