Question

Write the equation of the circle in standard form. Find the center, radius, intercepts, and graph the circle.
x^2+y^2-12x-4x+36=0

How on earth do I do this?

Answer 1

I meant to say *4y instead of 4x

Answer 2

x^2+y^2-12x-4y+36=0

you will have to complete the square and put the equation into standard form

x^2 - 12x + .... + y^2 - 4y + .... = -36 + .... + ....
x^2 - 12x + 36 + y^2 - 4y + 4 = -36 + 36 + 4
(x-6)^2 + (y-2)^2 = 4

so we have a circle with centre at (6,2) and radius 2

for x-intercept, let y = 0
(x-6)^2 + 4 = 4
x-6 = 0
x = 6
The circle touched the x-axis at (6,0)

for the y-intercept, let x = 0
36 + (y-2)^2 = 4
(y-2)^2 = -32
this has no solution, we can't take the √ of a
No y-intercepts

Your sketch will confirm that.