Question

A circle has a center that falls on the line `y = 5/4x +8` and passes through `( 4 ,7 )` and `(2 ,5 )`. What is the equation of the circle?

Answer 1

center is `-=(4/9,8 5/9)`
`(x-4/9)^2+y-8 5/9)^2=(3 5/9)^2+(1 5/9)^2`
or
`-=(0.444,8.555)`
#(x-0.444)^2+(y-8.555)^2=15.06

Explanation:

Let
`(x,y)`
be the center of the circle
`(4,7)`
and
`(2,5)`
lie on the perimeter.
`(4,7) and (2,5)`
are equidistant from (x,y)
`(x-4)^2+(y-7)^2=(x-2)^2+(y-5)^2`
`x^2-8x+16+y^2-14y+49=x^2-4x+4+y^2-10y+25`
`-8x-14y+16+49=-4x-10y+4+25`
`-8x-14y+65=-4x-10y+29`
`8x-4x+14y-10y+29-65=0`
`4x+4y-36=0`
`x+y=9`
Center lies on the line
`y=5/4x+8`
`x+5/4x+8=9`
`9/4x=9-8`
`9/4x=1`
`x=4/9`
`x+y=9`
`4/9+y=9`
`4/9+y=9`
`y=9-4/9`
`y=8 5/9`
`y=5/4x+8`
`8 5/9=5/4(4/9)+8`
`8 5/9=5/9)+8`
`8 5/9=8 5/9`
lhs=rhs
`(x-4)^2+(y-7)^2=(x-2)^2+(y-5)^2`
`(4/9-4)^2+(8 5/9-7)^2=(4/9-2)^2+(8 5/9-5)^2`
`(3 5/9)^2+(1 5/9)^2=(1 5/9)^2+(3 5/9)^2`
lhs=rhs
and the radius is
`r=sqrt((3 5/9)^2+(1 5/9)^2)`
`r=3.88`
center is `-=(4/9,8 5/9)`

`(x-4/9)^2+y-8 5/9)^2=(3 5/9)^2+(1 5/9)^2`
`-=(0.444,8.555)`
#(x-0.444)^2+(y-8.555)^2=15.06