A circle has a center that falls on the line y = 2/3x +7 and passes through ( 3 ,4 ) and (6 ,4 ). What is the equation of the circle?

1 Answer

(x - 9/2)^2 + (y - 10)^2 = 153/4

Explanation:

Circle C: (x - a)^2 + (y - b)^2 = R^2

(a,b) in r: b = 2/3 a + 7

C: (x - a)^2 + (y - 2/3 a - 7)^2 = R^2

(3, 4) in C: (3 - a)^2 + (4 - 2/3 a - 7)^2 = R^2 // equation 1

(6, 4) in C: (6 - a)^2 + (4 - 2/3 a - 7)^2 = R^2 // equation 2

Subtract: (1) - (2)

(3 - a)^2 - (6 - a)^2 = 0

3 - a = 6 - a   or   3 - a = - (6 - a)

0a = 3   or   3 + 6 = a + a

a = 9/2

b = 2/3 * 9/2 + 7 = 10

(1) \Rightarrow R^2 = (3 - 9/2)^2 + (4 - 10)^2

R^2 = (3/2)^2 + 36 = (9 + 144)/4 = 153/4