How do you graph #X^2+Y^2-16X+4Y+52=0#?

1 Answer
Jan 14, 2017

Answer:

This is a circle of radius #4# with centre #(8, -2)#

Explanation:

Complete the square for both #x# and #y# in order to get this in the form of the standard equation of a circle:

Given:

#x^2+y^2-16x+4y+52=0#

Reorganise as:

#color(blue)(x^2-16x+64)+color(green)(y^2+4y+4)-16=0#

That is:

#color(blue)(x^2-2(8x)+8^2)+color(green)(y^2+2(2y)+2^2)-4^2=0#

Hence:

#color(blue)((x-8)^2)+color(green)((y+2)^2) = 4^2#

which is (more or less) in the form:

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

with #(h, k) = (8, -2)# being the centre of the circle and #r = 4# being the radius.

graph{(x^2+y^2-16x+4y+52)((x-8)^2+(y+2)^2-0.038)=0 [-4.08, 15.92, -6.84, 3.16]}