Question

The abscissae of the two points A and B are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are the roots of the equation $x^2+2px-q^2=0$. Find the equation of the circle with AB as diameter. Also, find its radius.

Answer 1

The given equations are
$x^2+2ax-b^2—0\dots \left(i\right)x^2+2px-q^2=0\dots \left(ii\right)$
We have, the roots of (i) will give the absissae and the roots of (ii) will give the ordinate of A B respectively.
Now roots of (i)
$x=\frac{-2a\pm \sqrt{4a^2+4b^2}}{2}=-9\pm \sqrt{a^2+4b^2}$
Roots of (ii)
$x=\frac{-2p\pm \sqrt{4p^2+4q^2}}{2}=-p\pm \sqrt{p^2+q^2}$
Coordinates of
$A=(-a+\sqrt{a^2+b^2},-p\sqrt{p^2+q^2})B=(-a-\sqrt{a^2+b^2},-p-\sqrt{p^2+q^2})$
So the equation of circle is
$(x+a-\sqrt{a^2+b^2})(a+a+\sqrt{a^2+b^2})+(y+p-\sqrt{p^2+q^2})(y+p+\sqrt{p^2+q^2})=0\Rightarrow x^2+y^2+2ax+2py-(a^2+b^2+p^2+q^2)+q^2+p^2=0\Rightarrow x^2+y^2+2ax+2py-(b^2+q^2)=0$
The radius is
$r=\sqrt{g^2+f^2-c}=\sqrt{a^2+p^2-(b^2+q^2)}=\sqrt{a^2+b^2+p^2+q^2}$