How do you graph # (x-2)^2 + (y+1)^2 = 13#?

1 Answer
Jan 29, 2016


Explained below


The given equation represents a circle centered at (2, -1) and radius equal to #sqrt13#. First locate the point (2, -1) on the graph and the next task is to measure a radius equal to #sqrt13#. For this draw a right triangle with sides 12 and 5 units. Then with radius equal to the hypotenuse draw the required circle using (2, -1) as the center using compass.