Enter the equation of the circle described below. center (3, -2), radius = 5

Solution:

Given,

Center (3, -2), radius = 5

We know that the equation of circle with centre (h,k) and radius r units is

(x - h)² + (y - k)² = r²

The equation of the circle with centre (3, -2), radius = 5 is

(x - 3)² + (y + 2)² = 5²

Using the algebraic identities

(a - b)² = a² + b² - 2ab

(a + b)² = a² + b² + 2ab

x² - 6x + 9 + y² +4y + 4 = 25

x² + y² - 6x + 4y + 13 = 25

So we get

x² + y² - 6x + 4y + 13 - 25 = 0

x² + y² - 6x + 4y - 12 = 0.

Therefore, the equation of the circle is x² + y² - 6x + 4y - 12 = 0.

Enter the equation of the circle described below. center (3, -2), radius = 5

Summary:

The equation of the circle with centre (3, -2), radius = 5 is x² + y² - 6x + 4y - 12 = 0.