Answer to Question #138699 in Analytic Geometry for Jessa Mae

Question

What is the equation of the circle with center at (2,5) passing through (-5,5) and (-1,1)

Answer 1

The equation of the circle of radius $R$ with center at $(a,b)$ is $(x-a)^2+(y-b)^2=R^2$. In our case, $(x-2)^2+(y-5)^2=R^2$. Since the circle passing through $(-5,5)$, we have $(-5-2)^2+(5-5)^2=R^2$, and consequently $R=7$. On the other hand, since the circle passing through $(-1,1)$, we have $(-1-2)^2+(1-5)^2=R^2$, and therefore $R=5$. This contradiction shows that such a circle does not exist.