Question

Find an equation of the circle that has center (-3,3) and passes through (6,-2)

Answer 1

Answer:

( x + 3)^2 + ( y - 3)^2 = 106

Step-by-step explanation:

The equation of the circle with a center and a point

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

( -3 , 3) - center - (a, b)

a = -3

b = 3

( 6 , -2) - point - ( x, y)

x = 6

y = -2

Step 1: substitute the center into the equation

( x -(-3)^2 + ( y - 3)^2 = r^2

( x + 3)^2 + ( y - 3)^2 = r^2

Step 2: sub the point into the equation

( 6 + 3)^2 + ( -2 - 3)^2 = r^2

9^2 + ( -5)^2 = r^2

81 + 25 = r^2

106 = r^2

Step 3: sub the radius into the equation

( x + 3)^2 + ( y - 3)^2 = 106

The equation of the circle is

( x + 3)^2 + ( y - 3)^2 = 106