A circle has a center that falls on the line y = 7/2x +3 and passes through (1 ,2 ) and (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

Explanation:

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

C=((1+8)/2,(5+2)/2)=(9/2,7/2)

The slope of AB is =(5-2)/(8-1)=(3)/(7)

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

The equation of the line passing trrough C and perpendicular to AB is

y-7/2=-7/3(x-9/2)

y=-7/3x+21/2+7/2=-7/3x+14

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

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

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

35/6x=11

x=66/35

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

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

The radius of the circle is

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

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

=11069/1225

The equation of the circle is

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

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]}