A circle has a center that falls on the line $y = 7/6x +1$ and passes through $(1 ,4 )$ and $(8 ,5 )$ . What is the equation of the circle?

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Answer 1 · 2016-07-16T03:24:10

$(7x-30)^2 +49(y-6)^2 = 725$

The equation of the circle is

$(x-x_c)^2 +(y-y_c)^2 = R^2$

where $(x_c,y_c)$ is the center and $R$ the radius of the circle.

To find the center note that since the points $(1,4)$ and $(8,5)$ are on the circle, we have

$(1-x_c)^2 +(4-y_c)^2 = (8-x_c)^2 +(5-y_c)^2 = R^2$

Rearranging, we get a linear equation obeyed by $(x_c,y_c)$

$7x_c+y_c=36$

However, we already know

$y_c = 7/6 x_c +1$

Solving this pair of linear simultaneous equations gives

$x = 30/7, y=6$

Substituting this back in the equation $(1-x_c)^2 +(4-y_c)^2 = R^2$ , we get $R^2 = 725/49$ .

A circle has a center that falls on the line y = 7/6x +1 y = 7/6x +1 and passes through (1 ,4 )(1 ,4 ) and (8 ,5 )(8 ,5 ). What is the equation of the circle?