Question

The center of a circle is located at (3, 8), and the circle has a radius that is 5 units long. What is the general form of the equation for the circle?

Answer 1

The answer is : x2 + y2 − 6x − 16y + 48 = 0

Explanation:

I got it right on the Edmentum test.

Answer 2

The general equation of a circle is x² + y² - 6x - 16y + 48 = 0 .

Explanation:

The standard form of equation of circle.

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

Where (h,k) is the centre of a circle and r is the radius.

As given

The center of a circle is located at (3, 8), and the circle has a radius that is 5 units .

Put in the formula

`(x - 3)^(2) + (y - 8)^(2) = 5^(2)`

(By using the formula (a - b)² = a² + b²- 2ab)

As 5² = 25

x² + 9 - 6x + y² + 64 - 16y = 25

x² + y² - 6x - 16y + 9 + 64 = 25

x² + y² - 6x - 16y + 73 = 25

x² + y² - 6x - 16y + 73 - 25 = 0

x² + y² - 6x - 16y + 73 - 25 = 0

x² + y² - 6x - 16y + 48 = 0

(As the general equation of a circle are in the form x² + y² + Ax +By + F =0 Where A,B and C are constant .)

Therefore the general equation of a circle is x² + y² - 6x - 16y + 48 = 0 .