A circle has a center that falls on the line y = 2/9x +8 and passes through ( 2 ,5 ) and (1 ,7 ). What is the equation of the circle?

1 Answer
P dilip_k
Apr 15, 2016

Explanation:

Let the coordinate of the centre of the circle be (a,b) and radius is r.
Then its equation will be
(x-a)^2+(y-b)^2=r^2

The circle passes through (2,5) and (1,7) show these points will satisfy the equation of the circle, hence

(2-a)^2+(5-b)^2=r^2.......(1)
(1-a)^2+(7-b)^2=r^2.......(2)
Subtracting (2) from (1) we have

(2-a)^2-(1-a)^2+(5-b)^2-(7-b)^2=0

=>(2-a-1+a)(3-2a)+(5-b-7+b)(12-2b)=0
=>3-2a-24+4b=0
=>2a-4b=-21......(3)

Again The center (a,b) lies on the given line y=2/9x+8 So we can write b=2/9a+8......(4)

From (3) and (4)
we have
=>2a-4(2/9a+8)=-21

=>2a-8/9a-32=-21
=>2a-8/9a=11
=>10a=99
:,a=9.9
From (4)
b=2/9a+8
=>b=2/9xx9.9+8=10.2
From eq (1)
(2-9.9)^2+(5-10.2)^2=r^2.......(1)
=>r^2=62.41+27.04=89.45

Equation of the Circle
(x-9.9)^2+(y-10.2)^2=89.45^2

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