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

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Answer 1 · 2016-10-19T21:59:13

$(x - 38/3)^2 + (y - 142/3)^2 = (sqrt(19721)/3)^2$

The standard form for the equation of a circle is:

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

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

Using the two points, we write two equations:

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

Using the pattern $(a - b)^2 = a^2 - 2ab + b^2$ we expand the squares:

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

Subtract the first equation from the second:

$35 -10h - 3 + 2k = 0$

$32 -10h + 2k = 0$

$k = 5h - 16$

Substitute $(h, k)$ into the given equation:

$k = 7/2h + 3$

Set the right sides equal:

$5h - 16 = 7/2h + 3$

$10h - 32 = 7h + 6$

$h = 38/3$
$k = 142/3$

Substitute the center into the first equation:

$(1 - 38/3)^2 + (2 - 142/3)^2 = r^2$

$r = sqrt(19721)/3$

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