Question

The center of a circle is located at (6, -1). The radius of the circle is 4.

What is the equation of the circle in general form?

A. x^2 + y^2 - 12x + 2y + 33 = 0

B. x^2 + y^2 + 12x - 2y + 33 = 0

C. x^2 + y^2 - 12x + 2y + 21 = 0

D. x^2 + y^2 + 12x - 2y + 21 = 0

Answer 1

x^2+y^2+12x-2y+33=0

Explanation:

Answer 2

C)
`x^(2) +  y^(2) - 12x + 2y + 21 = 0`

Equation of a circle:

`(x-a)^(2) +(y-b)^(2) =r^(2)`

Substitute into the equation

a = 6

b = -1

r = 4

`(x-6)^(2) +(y--1)^(2) =4^(2)`

`(x-6)^(2) +(y+1)^(2) =16`

Expand and Simplify

`(x-6)^(2) = x^(2) - 12x + 36`

`(y+1)^(2) = y^(2) + 2y + 1`

`x^(2) - 12x + 36 + y^(2) + 2y + 1 = 16`

`x^(2) + y^(2) - 12x + 2y + 36 +  1 = 16`

`x^(2) + y^(2) - 12x + 2y + 36 +  1 - 16 = 0`

`x^(2) + y^(2) - 12x + 2y + 21 = 0`