A circle has a center that falls on the line $y = \frac{5}{8} x + 1$ $y = 5/8x +1$ and passes through $(5, 2)$ $( 5 ,2 )$ and $(3, 4)$ $(3 ,4 )$ . What is the equation of the circle?

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Answer 1 · 2017-06-10T11:59:42

The equation of the circle is ${(x - \frac{16}{3})}^{2} + {(y - \frac{13}{3})}^{2} = \frac{50}{9}$ $(x-16/3)^2+(y-13/3)^2=50/9$

Let $C$ $C$ be the mid point of $A = (5, 2)$ $A=(5,2)$ and $B = (3, 4)$ $B=(3,4)$

$C = (\frac{5 + 3}{2}, \frac{2 + 4}{2}) = (4, 3)$ $C=((5+3)/2,(2+4)/2)=(4,3)$

The slope of $A B$ $AB$ is $= \frac{4 - 2}{3 - 5} = - \frac{2}{2} = - 1$ $=(4-2)/(3-5)=-2/2=-1$

The slope of the line perpendicular to $A B$ $AB$ is $= 1$ $=1$

The equation of the line passing trrough $C$ $C$ and perpendicular to $A B$ $AB$ is

$y - 3 = 1 (x - 4)$ $y-3=1(x-4)$

$y = x - 1$ $y=x-1$

The intersection of this line with the line $y = \frac{5}{8} x + 1$ $y=5/8x+1$ gives the center of the circle.

$\frac{5}{8} x + 1 = x - 1$ $5/8x+1=x-1$

$\frac{3}{8} x = 2$ $3/8x=2$

$x = \frac{16}{3}$ $x=16/3$

$y = \frac{16}{3} - 1 = \frac{13}{3}$ $y=16/3-1=13/3$

The center of the circle is $(\frac{16}{3}, \frac{13}{3})$ $(16/3,13/3)$

The radius of the circle is

$r^{2} = {(\frac{16}{3} - 5)}^{2} + {(\frac{13}{3} - 2)}^{2}$ $r^2=(16/3-5)^2+(13/3-2)^2$

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

$= \frac{50}{9}$ $=50/9$

The equation of the circle is

${(x - \frac{16}{3})}^{2} + {(y - \frac{13}{3})}^{2} = \frac{50}{9}$ $(x-16/3)^2+(y-13/3)^2=50/9$

graph{((x-16/3)^2+(y-13/3)^2-50/9)(y-5/8x-1)(y-x+1)=0 [0.604, 9.372, 1.62, 6.002]}

A circle has a center that falls on the line y=58x+1y = 5/8x +1 and passes through (5,2) ( 5 ,2 ) and (3,4)(3 ,4 ). What is the equation of the circle?