A circle has a center that falls on the line y = 7/2x +1 and passes through (1 ,4 ) and (8 ,5 ). What is the equation of the circle?

1 Answer
Oct 7, 2016

Generic eqn of a circle with centre (a,b) and radius r is is (x-a)^2+(y-b)^2=r^2

We are told the centre (a,b) lies on the line y=7/2x+1; so;
b=7/2a+1=>2b=7a+2 ---- Equation [1]

We are also told that the circle passes through (1,4); so
(1-a)^2+(4-b)^2=r^2 ---- Equation [2A]
:. 1-2a+a^2+16-8b+b^2=r^2
:. a^2+b^2 -2a-8b+17=r^2 ---- Equation [2]

We are also told that the circle passes through (8,5); so
(8-a)^2+(5-b)^2=r^2
:. 64-16a+a^2+25-10b+b^2=r^2
:. a^2+b^2 -16a-10b+89=r^2 ---- Equation [3]

Equation [3] - Equation [2] gives;
-16a-2a-10b-8b+89-17=0
:.-18a-18b+72=0
:.18a+18b=72
:.a+b=4=>b=4-a ---- Equation [4]

Substituting this value of b into Equation [1] gives:
2(4-a)=7a+2
:. 8-2a=7a+2
:. 9a=6
:. a=2/3 Substitute into Equation [4] gives b=4-2/3=10/3

We know know a=2/3 and b=10/3; so substitute these into Equation [2A] to find r:
(1-2/3)^2+(4-10/3)^2=r^2
:.(1/3)^2+(2/3)^2=r^2
:.1/9+4/9=r^2
:.r^2=5/9

So the required equation is (x-2/3)^2+(y-10/3)^2=5/9