write an equation for a circle that has a center at (-6 5) and passes through the point (-11 3)

Question

I dont understand

Answer 1

Equation of a circle with center (h,k) and radius r is (x-h)² + (y-k)² = r².

Since the center is (-6,5), we have (x+6)² + (y-5)² = r².

Since (-11,3) is on the circle, it must satisfy the equation of the circle. So, (-11+6)² + (3-5)² = r².

Therefore, r² = 29.

Equation of circle: (x+6)² + (y-5)² = 29.

Answer 2

Hello Ret,

The standard equation for a circle is:

(x-h)^2 + (y-k)^2 = r^2

Where the center of the circle is at (h,k) and r is the radius of the circle. Well, we already know the center of the circle, so we can plug it into the equation to get:

(x + 6)^2 + (y - 5)^2 = r^2

To find the radius, we need to find the distance from the center of the circle to a point on the circle. They give us a point on the circle so, by using the distance formula:

d = root[(-11 + 6)^2 + (3 - 5)^2]

d = root[25 + 4]

d = root[29] ; therefore r = root[29] and r^2 = 29.

Plug into the equation to get

(x + 6)^2 + (y - 5)^2 = 29