A circle has a center that falls on the line $y = \frac{11}{7} x + 8$ $y = 11/7x +8$ and passes through $(9, 1)$ $( 9 ,1 )$ and $(8, 7)$ $(8 ,7 )$ . What is the equation of the circle?

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A circle has a center that falls on the line $y = \frac{11}{7} x + 8$ $y = 11/7x +8$ and passes through $(9, 1)$ $( 9 ,1 )$ and $(8, 7)$ $(8 ,7 )$ . What is the equation of the circle?

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Answer 1 · 2017-05-08T04:01:39

${(x + 3.8559)}^{2} + {(y - 1.9407)}^{2} = {(12.8903)}^{2}$ $(x+3.8559)^2+(y-1.9407)^2=(12.8903)^2$

Explanation:

We will make two equations with two variables, where $(x, y)$ $(x,y)$ is the center of the circle.

The first equation is the given line, which passes through the center of the circle. Isolate $x$ $x$ so it's more convenient to substitute:
$y = \frac{11}{7} x + 8$ $y=11/7x+8$
$y - 8 = \frac{11}{7} x$ $y-8=11/7x$
$x = \frac{7}{11} y - \frac{56}{11}$ $x=7/11y-56/11$

We know that the points $(9, 1)$ $(9,1)$ and $(8, 7)$ $(8,7)$ are an equal distance away from the center, so use the distance formula (direct variant of Pythagorean theorem) to determine the center and radius:

$Distance = \sqrt{{(y_{2} - y_{1})}_{2}^{2} + {(x_{2} - x_{1})}_{2}^{2}}$ $"Distance "=sqrt((y_2-y_1)^2+(x_2-x_1)^2)$

$\sqrt{{(\frac{11}{7} x + 8 - 7)}^{2} + {(\frac{7}{11} y - \frac{56}{11} - 8)}^{2}} = \sqrt{{(\frac{11}{7} x + 8 - 1)}^{2} + {(\frac{7}{11} y - \frac{56}{11} - 9)}^{2}}$ $sqrt[(11/7x+8-color(blue)(7))^2+(7/11y-56/11-color(blue)(8))^2]=sqrt[(11/7x+8-color(blue)(1))^2+(7/11y-56/11-color(blue)(9))^2]$

Substitute $y = \frac{11}{7} x + 8$ $y=11/7x+8$ into the equation above:

$\sqrt{{(\frac{11}{7} x + 8 - 7)}^{2} + {(\frac{7}{11} (\frac{11}{7} x + 8) - \frac{56}{11} - 8)}^{2}} = \sqrt{{(\frac{11}{7} x + 8 - 1)}^{2} + {(\frac{7}{11} (\frac{11}{7} x + 8) - \frac{56}{11} - 9)}^{2}}$ $sqrt[(11/7x+8-color(blue)(7))^2+(7/11 color(red)((11/7x+8)) -56/11-color(blue)(8))^2]=sqrt[(11/7x+8-color(blue)(1))^2+(7/11 color(red)((11/7x+8)) -56/11-color(blue)(9))^2]$

$\sqrt{{(\frac{11}{7} x + 1)}^{2} + {(x - 8)}^{2}} = \sqrt{{(\frac{11}{7} x + 7)}^{2} + {(x - 9)}^{2}}$ $sqrt[(11/7x+1)^2+(x-8)^2]=sqrt[(11/7x+7)^2+(x-9)^2]$

The exact answers can be found by squaring each side, then expanding the polynomials to get a quadratic equation, but only one of the values for $x$ $x$ will work when substituted back in.

(Using a graphing calculator)
$x \approx - 3.8559$ $color(green)(x approx -3.8559)$

$y \approx 1.9407$ $color(green)(y approx 1.9407$

$Radius \approx 12.8903$ $color(green)("Radius"approx 12.8903$ (distance from center to either one of the points)

Now plug these in to the standard form for the equation of a circle to get:
${(x + 3.8559)}^{2} + {(y - 1.9407)}^{2} = {(12.8903)}^{2}$ $(x+3.8559)^2+(y-1.9407)^2=(12.8903)^2$

Answer 2 · 2017-05-08T06:55:22

${(x + \frac{455}{118})}^{2} + {(y - \frac{229}{118})}^{2} = {(9 + \frac{455}{118})}^{2} + {(1 - \frac{229}{118})}^{2}$ $(x+455/118)^2+(y-229/118)^2=(9+455/118)^2+(1-229/118)^2$

Explanation:

Centre falls on $y = \frac{11}{7} x + 8$ $y=11/7x+8$ .

as points $(9, 1)$ $(9,1)$ and $(8, 7)$ $(8,7)$ also fall on circle, centre falls on perpendicular bisector of segment joining $(9, 1)$ $(9,1)$ and $(8, 7)$ $(8,7)$ .

Observe that the slope of this segment is $\frac{7 - 1}{8 - 9} = - 6$ $(7-1)/(8-9)=-6$ . Hence slope of peependicular bisector wiuld be $\frac{1}{6}$ $1/6$ .

As their midpoint point is $(\frac{9 + 8}{2}, \frac{7 + 1}{2})$ $((9+8)/2,(7+1)/2)$ i.e. $(\frac{17}{2}, 4)$ $(17/2,4)$ , equation of perpendicular bisector is $y = \frac{1}{6} (x - \frac{17}{2}) + 4 = \frac{1}{6} x - \frac{17}{12} + 4 = \frac{1}{6} x + \frac{31}{12}$ $y=1/6(x-17/2)+4=1/6x-17/12+4=1/6x+31/12$

and centre is point of intersection of $y = \frac{1}{6} x + \frac{31}{12}$ $y=1/6x+31/12$ and $y = \frac{11}{7} x + 8$ $y=11/7x+8$

i.e. $\frac{11}{7} x + 8 = \frac{1}{6} x + \frac{31}{12}$ $11/7x+8=1/6x+31/12$ or $(\frac{11}{7} - \frac{1}{6}) x = \frac{31}{12} - 8$ $(11/7-1/6)x=31/12-8$

or $\frac{59}{42} x = - \frac{65}{12}$ $59/42x=-65/12$ i.e. $x = - \frac{65}{12} \times \frac{42}{59} = - \frac{1365}{354} = - \frac{455}{118}$ $x=-65/12xx42/59=-1365/354=-455/118$

and $y = \frac{1}{6} \times (- \frac{455}{118}) + \frac{31}{12} = - \frac{455}{708} + \frac{31}{12} = \frac{- 455 + 1829}{708} = \frac{1374}{708} = \frac{229}{118}$ $y=1/6xx(-455/118)+31/12=-455/708+31/12=(-455+1829)/708=1374/708=229/118$

and centre is $(- \frac{455}{118}, \frac{229}{118})$ $(-455/118,229/118)$ and radius is its distance from say $(9, 1)$ $(9,1)$ i.e.

$\sqrt{{(9 + \frac{455}{118})}^{2} + {(1 - \frac{229}{118})}^{2}}$ $sqrt((9+455/118)^2+(1-229/118)^2)$

and equation of circle is

${(x + \frac{455}{118})}^{2} + {(y - \frac{229}{118})}^{2} = {(9 + \frac{455}{118})}^{2} + {(1 - \frac{229}{118})}^{2}$ $(x+455/118)^2+(y-229/118)^2=(9+455/118)^2+(1-229/118)^2$

graph{((x+455/118)^2+(y-229/118)^2-(9+455/118)^2+(1-229/118)^2)(y-11/7x-8)((x-9)^2+(y-1)^2-0.04)((x-8)^2+(y-7)^2-0.04)(y-1/6x-31/12)=0 [-20.92, 19.08, -8.88, 11.12]}

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A circle has a center that falls on the line $y = \frac{11}{7} x + 8$ $y = 11/7x +8$ and passes through $(9, 1)$ $( 9 ,1 )$ and $(8, 7)$ $(8 ,7 )$ . What is the equation of the circle?

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A circle has a center that falls on the line y=117x+8y = 11/7x +8 and passes through (9,1) ( 9 ,1 ) and (8,7)(8 ,7 ). 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 $y = \frac{11}{7} x + 8$ $y = 11/7x +8$ and passes through $(9, 1)$ $( 9 ,1 )$ and $(8, 7)$ $(8 ,7 )$ . What is the equation of the circle?