Q.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 and the radius of the circle with AB as diameter.

The 'diameter form' of a circle, $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$, expands directly using the sum/product of roots of the two given quadratics.

Let the abscissae of $A,B$ be $x_1,x_2$ — the roots of $x^2+2ax-b^2=0$: $x_1+x_2=-2a,\ x_1x_2=-b^2$.

Let the ordinates of $A,B$ be $y_1,y_2$ — the roots of $x^2+2px-q^2=0$ (using $x$ as the dummy variable of that equation): $y_1+y_2=-2p,\ y_1y_2=-q^2$.

The circle with $AB$ as diameter has equation

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.$

Expanding using the symmetric sums:

$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=x^2+2ax-b^2,$

$(y-y_1)(y-y_2)=y^2-(y_1+y_2)y+y_1y_2=y^2+2py-q^2.$

Adding: …