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?

1 Answer
Shiva Prakash M V
Mar 7, 2018

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

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