We are given a set of four points: $A=(5,-3) $ $B=(2,1)$ $C=(-5,2)$ $D=(-9,-1)$ Find the equation of the circle circumscribed on the trapezoid.
My idea to solve this problem is to simply substitute these points to the equation of a circle and solve the system of equations. $$(5-x)^2+(-3-y)^2 =R^2$$ $$(2-x)^2 + (1-y)^2 =R^2$$ $$(-5-x)^2+(2-y)^2 = R^2$$ $$(-9-x)^2+(-1-y)^2=R^2 $$ However, this system only has three variables, thus one point is unnecessary. Apart from that, this will be a long and tedious calculation. Could you suggest me a better way to tackle this?