A circle has a center that falls on the line $y = \frac{3}{2} x + 1$ $y = 3/2x +1$ and passes through $(1, 2)$ $(1 ,2 )$ 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 $y = \frac{3}{2} x + 1$ $y = 3/2x +1$ and passes through $(1, 2)$ $(1 ,2 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?

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Answer 1 · 2016-11-08T18:00:12

The equation of the circle is:
${(x - \frac{34}{7})}^{2} + {(y - \frac{58}{7})}^{2} = {(\frac{\sqrt{2665}}{7})}^{2}$ $(x - 34/7)^2 + (y - 58/7)^2 = (sqrt2665/7)^2$

Explanation:

The equation of a circle is:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x - h)^2 + (y - k)^2 = r^2$

where $(h, k)$ $(h,k)$ is the center and r is the radius.

Substitute the point (h,k) into the equation for the line on which the center lies:

$k = \frac{3}{2} h + 1$ $k = 3/2h + 1$ $[1]$ $" [1]"$

We can use the two given points and the equation of a circle to write two more equations:

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

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

We can, temporarily, eliminate the variable r by setting the left side of equation [2] equal to the left side of equation [3]:

${(1 - h)}^{2} + {(2 - k)}^{2} = {(6 - h)}^{2} + {(1 - k)}^{2}$ $(1 - h)^2 + (2 - k)^2 = (6 - h)^2 + (1 - k)^2$

Use the pattern ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a - b)^2 = a^2 - 2ab + b^2$ , to expand the squares:

$1 - 2 h + h^{2} + 4 - 4 k + k^{2} = 36 - 12 h + h^{2} + 1 - 2 k + k^{2}$ $1 - 2h + h^2 + 4 - 4k + k^2 = 36 - 12h + h^2 + 1 - 2k + k^2$

The square terms are the same on both sides of the equation, therefore, they cancel:

$1 - 2 h + 4 - 4 k = 36 - 12 h + 1 - 2 k$ $1 - 2h + 4 - 4k = 36 - 12h + 1 - 2k$

Collect the 4 constant terms into a single term on the right:

$- 2 h - 4 k = 32 - 12 h - 2 k$ $-2h - 4k = 32 - 12h - 2k$

Collect the 2 h terms into a single term on the right:

$- 4 k = 32 - 10 h - 2 k$ $-4k = 32 - 10h - 2k$

Collect the 2 k terms into a single term on the left:

$- 2 k = 32 - 10 h$ $-2k = 32 - 10h$

Divide both sides by -2:

$k = 5 h - 16$ $k = 5h - 16$ $[4]$ $" [4]"$

Subtract equation [4] from equation [1]:

$k - k = \frac{3}{2} h - 5 h + 1 + 16$ $k - k = 3/2h - 5h + 1 + 16$

$0 = - \frac{7}{2} h + 17$ $0 = -7/2h + 17$

$h = \frac{34}{7}$ $h = 34/7$

Substitute 34/7 for h in equation [1]:

$k = \frac{3}{2} (\frac{34}{7}) + 1$ $k = 3/2(34/7) + 1$

$k = \frac{58}{7}$ $k = 58/7$

Substitute the values for h and k into equation [2] or [3] (I will use equation [2]):

${(1 - \frac{34}{7})}^{2} + {(2 - \frac{58}{7})}^{2} = r^{2}$ $(1 - 34/7)^2 + (2 - 58/7)^2 = r^2$

${(- \frac{27}{7})}^{2} + {(- \frac{44}{7})}^{2} = r^{2}$ $(-27/7)^2 + (-44/7)^2 = r^2$

$r^{2} = \frac{2665}{7^{2}}$ $r^2 = 2665/7^2$

$r = \frac{\sqrt{2665}}{7}$ $r = sqrt(2665)/7$

The equation of the circle is:
${(x - \frac{34}{7})}^{2} + {(y - \frac{58}{7})}^{2} = {(\frac{\sqrt{2665}}{7})}^{2}$ $(x - 34/7)^2 + (y - 58/7)^2 = (sqrt2665/7)^2$

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A circle has a center that falls on the line $y = \frac{3}{2} x + 1$ $y = 3/2x +1$ and passes through $(1, 2)$ $(1 ,2 )$ 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 = 3/2x +1 y=32x+1y = 3/2x +1 and passes through (1 ,2 )(1,2)(1 ,2 ) and (6 ,1 )(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 $y = \frac{3}{2} x + 1$ $y = 3/2x +1$ and passes through $(1, 2)$ $(1 ,2 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?