Question

Find the equation of the circle the end points of whose diameter are the centres of the circles x² + y² + 6x − 14y − 1 = 0 and x² + y² − 4x + 10y − 2 = 0.

Answer 1

Given:
$x^2+y^2+6x-14y-1=0$ ...(1)
And, $x^2+y^2-4x+10y-2=0$ ...(2)

Equations (1) and (2) can be rewritten as follows:
${\left(x+3\right)}^2+{\left(y-7\right)}^2=59$
And, ${\left(x-2\right)}^2+{\left(y+5\right)}^2=31$

Thus, the centres of the circles are (−3, 7) and (2, −5).

Hence, the equation of the circle, the end points of whose diameter are the centres of the given circles, is$\left(x+3\right)\left(x-2\right)+\left(y-7\right)\left(y+5\right)=0$, i.e. $x^2+y^2+x-2y-41=0$.