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 $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?

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A circle has a center that falls on the line $y = \frac{2}{3} x + 7$ $y = 2/3x +7$ and passes through $(3, 4)$ $( 3 ,4 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?

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Answer 1 · 2016-12-03T19:43:00

${(x - 27)}^{2} + {(y - 25)}^{2} = {(3 \sqrt{113})}^{2}$ $(x - 27)^2 + (y - 25)^2 = (3sqrt(113))^2$

Explanation:

The general form of 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, and r is the radius.

Use equation [1] to write two equations, using the given points:

${(3 - h)}^{2} + {(4 - k)}^{2} = r^{2} [2]$ $(3 - h)^2 + (4 - k)^2 = r^2" [2]"$
${(6 - h)}^{2} + {(1 - k)}^{2} = r^{2} [3]$ $(6 - h)^2 + (1 - k)^2 = r^2" [3]"$

Temporarily eliminate r by setting the left side of equation [2] equal to the left side of equation [3]:

${(3 - h)}^{2} + {(4 - k)}^{2} = {(6 - h)}^{2} + {(1 - k)}^{2}$ $(3 - h)^2 + (4 - k)^2 = (6 - h)^2 + (1 - k)^2$

Expand the squares, using the pattern ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a - b)^2 = a^2 - 2ab + b^2$ :

$9 - 6 h + h^{2} + 16 - 8 k + k^{2} = 36 - 12 h + h^{2} + 1 - 2 k + k^{2}$ $9 - 6h + h^2 + 16 - 8k + k^2 = 36 - 12h + h^2 + 1 - 2k + k^2$

The square terms cancel:

$9 - 6 h + 16 - 8 k = 36 - 12 h + 1 - 2 k$ $9 - 6h + 16 - 8k = 36 - 12h + 1 - 2k$

Collect all of the constant terms into a single term on the right:

$- 6 h - 8 k = - 12 h - 2 k + 12$ $-6h - 8k = -12h - 2k + 12$

Collect all of the h terms into a single term on the right:

$- 8 k = - 6 h - 2 k + 12$ $- 8k = -6h - 2k + 12$

Collect all of the k terms into a single term on the left:

$- 6 k = - 6 h + 12$ $-6k = -6h + 12$

Divide both sides of the equation by -6:

$k = h - 2 [4]$ $k = h - 2" [4]"$

Evaluate the given equation at the point $(h, k)$ $(h, k)$

$k = \frac{2}{3} h + 7 [5]$ $k = 2/3h + 7" [5]"$

Set the right side of equation [4] equal to the right side of equation [5]:

$h - 2 = \frac{2}{3} h + 7$ $h - 2 = 2/3h + 7$

$\frac{1}{3} h = 9$ $1/3h = 9$

$h = 27$ $h = 27$

Substitute 27 for h in equation [5]:

$k = \frac{2}{3} (27) + 7$ $k = 2/3(27) + 7$

$k = 18 + 7$ $k = 18 + 7$

$k = 25$ $k = 25$

Substitute the center, $(27, 25)$ $(27,25)$ , into equation [2] and solve for r:

${(3 - 27)}^{2} + {(4 - 25)}^{2} = r^{2}$ $(3 - 27)^2 + (4 - 25)^2 = r^2$

${(- 24)}^{2} + {(- 21)}^{2} = r^{2}$ $(-24)^2 + (-21)^2 = r^2$

$r^{2} = 1017$ $r^2 = 1017$

$r = \sqrt{1017} = 3 \sqrt{113}$ $r = sqrt(1017) = 3sqrt(113)$

Substitute the center, $(27, 25)$ $(27,25)$ , and the radius, $3 \sqrt{113}$ $3sqrt(113)$ into equation [1]:

${(x - 27)}^{2} + {(y - 25)}^{2} = {(3 \sqrt{113})}^{2}$ $(x - 27)^2 + (y - 25)^2 = (3sqrt(113))^2$

This 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 $(3, 4)$ $( 3 ,4 )$ and $(6, 1)$ $(6 ,1 )$ . What is the equation of the circle?

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Explanation:

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