A circle has a center that falls on the line $y = \frac{5}{2} x + 1$ $y = 5/2x +1$ and passes through $(8, 2)$ $(8 ,2 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?

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A circle has a center that falls on the line $y = \frac{5}{2} x + 1$ $y = 5/2x +1$ and passes through $(8, 2)$ $(8 ,2 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?

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Answer 1 · 2017-12-30T18:11:11

${(x - \frac{29}{9})}^{2} + {(y - \frac{163}{18})}^{2} = {(5 \frac{\sqrt{941}}{18})}^{2}$ $(x - 29/9)^2+(y-163/18)^2 = (5sqrt(941)/18)^2$

Explanation:

The standard Cartesian form for the equation of a circle is:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2} [1]$ $(x - h)^2+(y-k)^2 = r^2" [1]"$

where, $(x, y)$ $(x,y)$ is any point on the circle, $(h, k)$ $(h,k)$ is the center point and $r$ $r$ is the radius.

We may use equation [1] and the points, $(8, 2)$ $(8,2)$ and $(6, 1)$ $(6,1)$ to write 2 unique equations:

${(8 - h)}^{2} + {(2 - k)}^{2} = r^{2} [2]$ $(8 - h)^2+(2-k)^2 = r^2" [2]"$
${(6 - h)}^{2} + {(1 - k)}^{2} = r^{2} [3]$ $(6 - h)^2+(1-k)^2 = r^2" [3]"$

To obtain a third unique equation, we substitute the center point, $(h, k)$ $(h,k)$ , into the given linear equation:

$k = \frac{5}{2} h + 1 [4]$ $k = 5/2h+1" [4]"$

Expand the squares within equations [2] and [3]:

$64 - 16 h + h^{2} + 4 - 4 k + k^{2} = r^{2} [2.1]$ $64 - 16h+h^2+4-4k+k^2 = r^2" [2.1]"$
$36 - 12 h + h^{2} + 1 - 2 k + k^{2} = r^{2} [3.1]$ $36 - 12h+h^2+1-2k+k^2 = r^2" [3.1]"$

Subtract equation [2.1] from equation [3.1]:

$4 h + 2 k - 31 = 0 [5]$ $4h+2k -31=0" [5]"$

Substitute equation [4] into equation [5]:

$4 h + 2 (\frac{5}{2} h + 1) - 31 = 0$ $4h+2(5/2h+1) -31=0$

$4 h + 5 h + 2 - 31 = 0$ $4h+5h+2-31=0$

$h = \frac{29}{9}$ $h= 29/9$

Substitute this into equation [4], to obtain the value of k:

$k = \frac{5}{2} (\frac{29}{9}) + 1$ $k = 5/2(29/9)+1$

$k = \frac{163}{18}$ $k = 163/18$

One may substitute the values of h and k into either equation [2] or [3] to obtain the value of r; I shall use equation [2]:

${(8 - \frac{29}{9})}^{2} + {(2 - \frac{163}{18})}^{2} = r^{2}$ $(8 - 29/9)^2+(2-163/18)^2 = r^2$

${(\frac{144}{18} - \frac{58}{18})}^{2} + {(\frac{36}{18} - \frac{163}{18})}^{2} = r^{2}$ $(144/18 - 58/18)^2+(36/18-163/18)^2 = r^2$

${(\frac{86}{18})}^{2} + {(- \frac{127}{18})}^{2} = r^{2}$ $(86/18)^2+(-127/18)^2 = r^2$

$r = \frac{\sqrt{23525}}{18}$ $r= sqrt(23525)/18$

$r = 5 \frac{\sqrt{941}}{18}$ $r = 5sqrt(941)/18$

Substitute the values of $h, k, and r$ $h, k , and r$ into equation [1], to obtain the equation of the circle:

${(x - \frac{29}{9})}^{2} + {(y - \frac{163}{18})}^{2} = {(5 \frac{\sqrt{941}}{18})}^{2}$ $(x - 29/9)^2+(y-163/18)^2 = (5sqrt(941)/18)^2$

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A circle has a center that falls on the line $y = \frac{5}{2} x + 1$ $y = 5/2x +1$ and passes through $(8, 2)$ $(8 ,2 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?

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A circle has a center that falls on the line $y = \frac{5}{2} x + 1$ $y = 5/2x +1$ and passes through $(8, 2)$ $(8 ,2 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?