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

\[\left( x + 3 \right)\left( x - 2 \right) + \left( y - 7 \right)\left( y + 5 \right) = 0\]

\[x^2 + y^2 + x - 2y - 41 = 0\]