A circle has a center that falls on the line y = 6/7x +7  and passes through  ( 7 ,8 ) and (3 ,9 ). What is the equation of the circle?

Nov 8, 2016

graph{(6/7x+7-y)(y-17/2-4(x-5))(y-9+1/4(x-3))((x-7)^2+(y-8)^2-0.04)((x-3)^2+(y-9)^2-0.04)((x-259/44)^2+(y-265/22)^2-0.04)((x-259/44)^2+(y-265/22)^2-34085/44^2)=0 [-5, 19, 5, 17]}

${\left(x - \frac{259}{44}\right)}^{2} + {\left(y - \frac{265}{22}\right)}^{2} = \frac{34085}{44} ^ 2$

Explanation:

Call the points $A$ and $B$
Get the direction of $\overline{A B}$

${m}_{\overline{A B}} = \frac{8 - 9}{7} - 3 = - \frac{1}{4}$

The direction of $\overline{A B}$ axis is

${m}_{\overline{A B} \vdash} = 4$

Let $M$ be the medium point of $\overline{A B}$, then

M=((3+7)/2;(9+8)/2)=(5;17/2)

The equation of $\overline{A B}$ axis is

$y - \frac{17}{2} = 4 \left(x - 5\right)$

The center $C$ of the circle must beleng to $\overline{A B}$ axis so

$\left\{\begin{matrix}{y}_{C} = 4 \left({x}_{C} - 5\right) + \frac{17}{2} \\ {y}_{C} = \frac{6}{7} {x}_{C} + 7\end{matrix}\right.$

Solving the system we obtain

C=(259/44;265/22)

and for the radius

$r = \sqrt{{\left({x}_{A} - {x}_{C}\right)}^{2} + {\left({y}_{A} - {y}_{C}\right)}^{2}} = \sqrt{{\left(\frac{259}{44} - 3\right)}^{2} + {\left(\frac{265}{22} - 9\right)}^{2}} = \frac{\sqrt{34085}}{44}$