A circle has a center that falls on the line $y = 7/9x +7$ and passes through $( 7 ,3 )$ and $(5 ,1 )$ . What is the equation of the circle?

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Answer 1 · 2016-10-24T10:43:14

THe equation of the circle is $(x-11/16)^2+(y-117/16)^2=58.4$

Let the center of the circle be $a,b$

Then the equation of the circle is
$(x-a)^2+(y-b)^2=r^2$
Putting the points $(7,3)$ and $(5,1)$ in the equation of the circle

$(7-a)^2+(3-b)^2=r^2$ and $(5-a)^2+(1-b)^2=r^2$
so $(7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2$
Expanding the equations

$49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2$

$49-14a+9-6b=25-10a+1-2b$

$58-14a-6b=26-10a-2b$
$32=4a+4b$ $=>$ $8=a+b$ this is equation 1

From the equation of the line, we have $b=(7a)/9+1$
$9b=7a+9$ this is equation 2
Solving for a and b , we get
$a=11/16$ and $b=117/16$

Then we calculate $r$
$r^2=(7-11/16)^2+(3-117/16)^2$
$r^2=(101^2+69^2)/16$
$r=7.65$
So the equation of the circle is
$(x-11/16)^2+(y-117/16)^2=58.4$

A circle has a center that falls on the line y = 7/9x +7 y = 7/9x +7 and passes through ( 7 ,3 ) ( 7 ,3 ) and (5 ,1 )(5 ,1 ). What is the equation of the circle?