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

Question

1435 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-18T14:03:13

${(x - \frac{9}{2})}^{2} + {(y - 10)}^{2} = \frac{153}{4}$ $(x - 9/2)^2 + (y - 10)^2 = 153/4$

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

$(a, b) \in r : b = \frac{2}{3} a + 7$ $(a,b) in r: b = 2/3 a + 7$

$C : {(x - a)}^{2} + {(y - \frac{2}{3} a - 7)}^{2} = R^{2}$ $C: (x - a)^2 + (y - 2/3 a - 7)^2 = R^2$

$(3, 4) \in C : {(3 - a)}^{2} + {(4 - \frac{2}{3} a - 7)}^{2} = R^{2}$ $(3, 4) in C: (3 - a)^2 + (4 - 2/3 a - 7)^2 = R^2$ // equation 1

$(6, 4) \in C : {(6 - a)}^{2} + {(4 - \frac{2}{3} a - 7)}^{2} = R^{2}$ $(6, 4) in C: (6 - a)^2 + (4 - 2/3 a - 7)^2 = R^2$ // equation 2

Subtract: (1) - (2)

${(3 - a)}^{2} - {(6 - a)}^{2} = 0$ $(3 - a)^2 - (6 - a)^2 = 0$

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

$0 a = 3$ $0a = 3$ or $3 + 6 = a + a$ $3 + 6 = a + a$

$a = \frac{9}{2}$ $a = 9/2$

$b = \frac{2}{3} \cdot \frac{9}{2} + 7 = 10$ $b = 2/3 * 9/2 + 7 = 10$

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

$R^{2} = {(\frac{3}{2})}^{2} + 36 = \frac{9 + 144}{4} = \frac{153}{4}$ $R^2 = (3/2)^2 + 36 = (9 + 144)/4 = 153/4$

A circle has a center that falls on the line y = 2/3x +7 y=23x+7y = 2/3x +7 and passes through ( 3 ,4 )(3,4) ( 3 ,4 ) and (6 ,4 )(6,4)(6 ,4 ). What is the equation of the circle?