Question

Find the equation of the circle, in standard form whose center is (3,2) and passes through the point (6,5). (10 pts.) 2.

Answer 1

The equation of a circle in standard form is given by

(x - h)² + (y - k)² = r² where

(h, k) is the center and

'r' is the radius of circle

Now given that the center is (3, 2) implies (h, k) = (3, 2) implies

h = 3 and k = 2

Given that the circle passes through (6, 5)

radius is defined as the distance between center and any point on the circle

Now given center is (3, 2) and point on circle is (6, 5) implies

Radius is given by

r = √((6 - 3)² + (5 - 2)²

r = √(3² + 3²)

r = √(9 + 9)

r = √18

Therefore the equation of required circle in standard form is

(x - 3)² + (y - 2)² = (√18)² OR

(x - 3)² + (y - 2)² = 18