Question

The abscissae of the two points A and B are the roots of the equation x² + 2ax − b² = 0 and their ordinates are the roots of the equation x² + 2px − q² = 0. Find the equation of the circle with AB as diameter. Also, find its radius.

Answer 1

Roots of equation x² + 2ax − b² = 0 are

\[- a \pm \sqrt{a^2 + b^2}\] 

Roots of equation x² + 2px − q² = 0 are

\[- p \pm \sqrt{p^2 + q^2}\] .

Therefore, coordinates of A and B are

\[\left( - a + \sqrt{a^2 + b^2}, - p + \sqrt{p^2 + q^2} \right) and \left( - a - \sqrt{a^2 + b^2}, - p - \sqrt{p^2 + q^2} \right)\]respectively.

Hence, equation of circle is \[\left( x + a - \sqrt{a^2 + b^2} \right)\left( x + a + \sqrt{a^2 + b^2} \right) + \left( y + p - \sqrt{p^2 + q^2} \right)\left( y + p + \sqrt{p^2 + q^2} \right) = 0\]

\[\Rightarrow \left( x + a \right)^2 - a^2 - b^2 + \left( y + p \right)^2 - p^2 - q^2 = 0\]

\[\Rightarrow x^2 + y^2 + 2ax + 2yp - p^2 - q^2 = 0\]

Also, radius of circle is

\[\sqrt{a^2 + b^2 + p^2 + q^2}\]