A circle has a center that falls on the line y = 2/3x +7 and passes through (4 ,7 ) and (1 ,2 ). What is the equation of the circle?

1 Answer
Narad T.
Nov 9, 2016

The equation of the circle is (x+15/19)^2+(y-123/19)^2=23.2

Explanation:

Let the center of the circle be (a,b) and radius r
Then b=2/3a+7
The equation of the circle is (x-a)^2+(y-b)^2=r^2
Substituting the values of the 2 points
(4-a)^2+(7-b)^2=r^2
and (1-a)^2+(2-b)^2=r^2
:.(4-a)^2+(7-b)^2=(1-a)^2+(2-b)^2
16-8a+a^2+49-14b+b^2=1-2a+a^2+4-4b+b^2

65-8a-14b=5-2a-4b
6a+10b=60=>3a+5b=30
and the first equation is 3b-2a=21
Solving for a and b, we get (-15/19,123/19) as the center of the circle
we calculate the radius of the circle
r^2=(34/19)^2+(85/19)^2=23.21
r=4.82

Equation of the circle
(x+15/19)^2+(y-123/19)^2=23.2

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