Question

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

Answer 1

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]}