The equation of the circle, the end points of whose diameter are the centres of the circles $x^2+y^2+6 x-14…
The equation of the circle, the end points of whose diameter are the centres of the circles $x^2+y^2+6 x-14 y+5=0$ and $x^2+y^2-4 x+10 y-4=0$ is
$x^2+y^2-4 x+10 y-4=0$ is
$x^2+y^2-x-2 y+41=0$
$x^2+y^2+x-2 y-41=0$
$x^2+y^2+x-2 y-41=0$
$x^2+y^2-x+2 y-41=0$
Solution
Center of circle $x^2+y^2+6 x-14 y+5=0$ is $(-3,7)$ and centre of circle $x^2+y^2-4 x+10 y-4=0$ is $(2,-5)$
$\therefore \quad$ Equation of the required circle is
$\begin{aligned}
& (x+3)(x-2)+(y-7)(y+5)=0 \\
& \Rightarrow x^2+y^2+x-2 y-41=0
\end{aligned}$