A circle has a center that falls on the line $y = \frac{2}{3} x + 7$ $y = 2/3x +7$ and passes through $(5, 7)$ $(5 ,7 )$ and $(1, 2)$ $(1 ,2 )$ . What is the equation of the circle?

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A circle has a center that falls on the line $y = \frac{2}{3} x + 7$ $y = 2/3x +7$ and passes through $(5, 7)$ $(5 ,7 )$ and $(1, 2)$ $(1 ,2 )$ . What is the equation of the circle?

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Answer 1 · 2018-06-02T06:54:58

see below

Explanation:

the centre of the circle is on the intersection point between the assigned line and the line axis of the segment that joint the two points
straight for two points:
$\frac{y - y_{a}}{y_{b} - y_{a}} = \frac{x - x_{a}}{x_{b} - x_{a}}$ $(y-y_a)/(y_b-y_a) = (x-x_a)/(x_b-x_a)$
$\frac{y - 7}{2 - 7} = \frac{x - 5}{1 - 5}$ $(y-7)/(2-7) = (x-5)/(1-5)$
$\frac{y - 7}{- 5} = \frac{x - 5}{- 4}$ $(y-7)/-5 = (x-5)/-4$
$\frac{y - 7}{- 5} = \frac{x - 5}{- 4}$ $(y-7)/-5 = (x-5)/-4$
$- 4 (y - 7) = - 5 (x - 5))$ $-4 (y-7) = -5 (x-5))$
$- 4 y + 28 = - 5 x + 25$ $-4 y +28 = -5 x+ 25$
$- 4 y = - 5 x - 3$ $-4 y = -5 x-3$
$y = \frac{5}{4} x + \frac{3}{4}$ $y= 5/4 x + 3/4$

middle point M
$X_{m} = \frac{X_{b} + X_{a}}{2} = \frac{5 + 1}{2} = 3$ $X_m =( X_b+X_a) /2=(5+1)/2=3$
$Y_{m} = \frac{Y_{b} + Y_{a}}{2} = \frac{7 + 2}{2} = \frac{9}{2}$ $Y_m =( Y_b+Y_a) /2=(7+2)/2=9/2$

perpendicular line (m =-1/m') passing through M
$y - y_{M} = m (x - x_{m})$ $y-y_M= m (x-x_m)$
$y - \frac{9}{2} = - \frac{4}{5} (x - 3)$ $y-9/2= -4/5(x-3)$
$y = - \frac{4}{5} x + \frac{12}{5} + \frac{9}{2}$ $y= -4/5 x +12/5+9/2$
$y = - \frac{4}{5} x + \frac{59}{10}$ $y= -4/5 x +59/10$

making the system and solving with comparison
$\frac{2}{3} x + 7 = - \frac{4}{5} x + \frac{69}{10}$ $2/3 x +7= -4/5 x +69/10$
$\frac{22}{15} x = - \frac{1}{10}$ $22/15 x =-1/10$

$X_{C} = - \frac{3}{22}$ $X_C=-3/22$
$Y_{C} = - \frac{1}{11} + 7 = \frac{76}{11}$ $Y_C=-1/11+7=76/11$

$R = \sqrt{{(X_{c} - X_{a})}_{c}^{2} + {(Y_{c} - Y_{a})}_{c}^{2}} = \sqrt{{(- \frac{3}{22} - 1)}^{2} + {(\frac{76}{11} - 2)}^{2}} =$ $R=sqrt((X_c-X_a)^2 + (Y_c-Y_a)^2)=sqrt((-3/22-1)^2 + (76/11-2)^2)=$ ...

i have no mere time....

equation of circle
${(x - x_{c})}_{c}^{2} + {(y - y_{c})}_{c}^{2} = R^{2}$ $(x-x_c)^2 + (y-y_c)^2 = R^2$

Answer 2 · 2018-07-03T12:30:00

${(x + \frac{3}{44})}^{2} + {(y - \frac{153}{22})}^{2} = \frac{49733}{1936}$ $(x+3/44)^2+(y-153/22)^2=49733/1936$ .

Explanation:

Let, $C (h, k)$ $C(h,k)$ and $r$ $r$ be the centre and radius of the reqd. circle $S .$ $S.$

Since, $C$ $C$ is on the line $y = \frac{2}{3} x + 7, we have, k = \frac{2}{3} h + 7$ $y=2/3x+7," we have, "k=2/3h+7$ .

$:. C(h,k)=C(h,2/3h+7)$ .

The points $P(5,7) and Q(1,2) in S$ .

$:. CP^2=r^2=CQ^2$ .

$:.(h-5)^2+(2/3h+7-7)^2=(h-1)^2+(2/3h+7-2)^2$ .

$:.(h^2-10h+25)+4/9h^2=h^2-2h+1+4/9h^2+20/3h+25$ .

$:.-10h+2h-20/3h=1$ .

$:. h=-3/44$ .

$:. k=2/3h+7=2/3(-3/44)+7=153/22$ .

Finally, $r^2=(h-5)^2+(2/3h)^2=(-3/44-5)^2+(-1/22)^2$ .

$;. r^2=49733/1936$ .

$:. S : (x-h)^2+(y-k)^2=r^2, i.e.,$

$(x+3/44)^2+(y-153/22)^2=49733/1936$ .

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A circle has a center that falls on the line $y = \frac{2}{3} x + 7$ $y = 2/3x +7$ and passes through $(5, 7)$ $(5 ,7 )$ and $(1, 2)$ $(1 ,2 )$ . What is the equation of the circle?

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Explanation:

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A circle has a center that falls on the line y = 2/3x +7 y=23x+7y = 2/3x +7 and passes through (5 ,7 )(5,7)(5 ,7 ) and (1 ,2 )(1,2)(1 ,2 ). What is the equation of the circle?

2 Answers

Explanation:

Explanation:

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A circle has a center that falls on the line $y = \frac{2}{3} x + 7$ $y = 2/3x +7$ and passes through $(5, 7)$ $(5 ,7 )$ and $(1, 2)$ $(1 ,2 )$ . What is the equation of the circle?