Question

Find the equation of the circle, the end points of whose diameter are (2, -3) and (-2, 4).
Find its centre and radius.

Answer 1

We will use equation in diameter form.

$(x-2)(x+2)+(y+3)(y-4)=0x^2-4+y^2-y-12=0x^2+y^2-y-16=0\dots \left(1\right)2g=0,2f=1\dots$ [From (1)]
$g=0,f=-\frac{1}{2}$
Centre of the circle is $(0,\frac{1}{2})$
$r=\sqrt{0+\frac{1}{4}+16}r=\frac{\sqrt{65}}{2}$