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 $\left(-a+\sqrt{a^2+b^2},-p+\sqrt{p^2+q^2}\right)\mathrm{and}\left(-\mathrm{a}-\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2},-\mathrm{p}-\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}\right)$ 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 $\sqrt{a^2+b^2+p^2+q^2}$.