Find the equation of the circle with diameter AB where A and B are the points (-1,2) and (3,3) respectively?

##x^2+y^2-2x-5y+3=0##

Firstly, we can find the centre of the circle by finding the midpoint of ##AB##. Since it the midpoint or the centre ##(h,k)## cut AB with equal ratio;

##Centre (h,k) = ((-1+3)/2,(2+3)/2)##

##(h,k)=(1,5/2)##

Then we can find the radius , ##r## of the circle by using the equation;

##r^2=(x-h)^2+(y-k)^2##

##r^=sqrt((x-h)^2+(y-k)^2##

Substitute the coordinate ##(h,k)=(1,5/2)## and any of ##A## or ##B## coordinates into equation. In this calculation I choose ##B##.

##r^=sqrt((3-1)^2+(3-5/2)^2##

##r=sqrt(17/4)##=##sqrt17/2##

To find ##c##, we can use the equation ##c=h^2+k^2-r^2## and substitute ##(h,k)=(1,5/2)## and ##r=sqrt17/2## into it.

##c=(1)^2+(5/2)^2-(sqrt17/2)^2##

##c=3##

Then we know that the equation of circle is known as;

##x^2+y^2-2hx-2ky+c=0##

Substitute only ##(h,k)=(1,5/2)## and ##c=3## and we get;

##x^2+y^2-2(1)x-2(5/2)y+3=0##

##x^2+y^2-2x-5y+3=0##