A circle has a center that falls on the line y = 7/2x +3 y=72x+3 and passes through (1 ,2 )(1,2) and (8 ,5 )(8,5). What is the equation of the circle?

1 Answer
Jul 7, 2017

The equation of the circle is (x-66/35)^2+(y-48/5)^2=71717/1225(x6635)2+(y485)2=717171225

Explanation:

Let CC be the mid point of A=(1,2)A=(1,2) and B=(8,5)B=(8,5)

C=((1+8)/2,(5+2)/2)=(9/2,7/2)C=(1+82,5+22)=(92,72)

The slope of ABAB is =(5-2)/(8-1)=(3)/(7)=5281=37

The slope of the line perpendicular to ABAB is =-7/3=73

The equation of the line passing trrough CC and perpendicular to ABAB is

y-7/2=-7/3(x-9/2)y72=73(x92)

y=-7/3x+21/2+7/2=-7/3x+14y=73x+212+72=73x+14

The intersection of this line with the line y=7/2x+3y=72x+3 gives the center of the circle.

-7/3x+14=7/2x+373x+14=72x+3

7/2x+7/3x=14-372x+73x=143

35/6x=11356x=11

x=66/35x=6635

y=7/2*(66/35)+3=48/5y=72(6635)+3=485

The center of the circle is (66/35,48/5)(6635,485)

The radius of the circle is

r^2=(1-66/35)^2+(2-48/5)^2r2=(16635)2+(2485)2

=(-31/35)^2+(-38/5)^2=(3135)2+(385)2

=11069/1225=110691225

The equation of the circle is

(x-66/35)^2+(y-48/5)^2=71717/1225(x6635)2+(y485)2=717171225

graph{((x-66/35)^2+(y-48/5)^2-71717/1225)(y-7/2x-3)(y+7/3x-14)=0 [-23.05, 12.98, -0.97, 17.06]}