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?

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Answer 1 · 2017-01-18T14:03:13

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

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

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

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