A circle has a center that falls on the line y = 3x +4 and passes through (4 ,4 ) and (9 ,4 ). What is the equation of the circle?

1 Answer
Leland Adriano Alejandro
Mar 25, 2016

Equation of the circle
(x-13/2)^2+(y-47/2)^2=1546/4

Explanation:

Let the center (h,k)

From the given line y=3x+4
we can write k=3h+4 because (h, k) is on the line

Using the given points (4, 4) and (9, 4)

radius=radius

sqrt((4-h)^2+(4-k)^2)=sqrt((9-h)^2+(4-k)^2)

also

(4-h)^2+(4-k)^2=(9-h)^2+(4-k)^2

16-8h+h^2=81-18h+h^2

18h-8h=81-16
10h=65
h=13/2

Using now the equation k=3h+4

k=3(13/2)+4

k=39/2+4

k=(39+8)/2

k=47/2

the center of the circle
(h, k)=(13/2, 47/2)

compute the radius r now

r^2=(4-h)^2+(4-k)^2

r^2=(4-13/2)^2+(4-47/2)^2

r^2=((8-13)/2)^2+((8-47)/2)^2

r^2=(-5/2)^2+(-39/2)^2

r^2=1546/4

We can now obtain the equation of the circle

(x-h)^2+(y-k)^2=r^2

(x-13/2)^2+(y-47/2)^2=1546/4

God bless....I hope the explanation is useful.

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