A circle has a center that falls on the line y = 7/9x +7 and passes through ( 4 ,5 ) and (8 ,1 ). What is the equation of the circle?

1 Answer
Narad T. ยท Shwetank Mauria
Mar 17, 2017

The equation of the circle is (x-45)^2+(y-42)^2=3050

Explanation:

The mid point of (4,5) and (8,1) is (6,3)

The slope of the line through (4,5) and (8,1) is

m_1=(1-5)/(8-4)=-4/4=-1

The slope of the line perpendicular to the line through (4,5) and (8,1) is

m_2=1

The equation of the line through (6,3) with a slope of 1 is

y-3=1(x-6)

y=x-6+3=x-3

The center of the circle lies on the intersection of lines

y=7/9x+7

and

y=x-3

Therefore,

7/9x+7=x-3

x(1-7/9)=7+3=10

2/9x=10

x=10*9/2=45

and

y=45-3=42

So,

The center of the circle is C=(45,42)

The radius is

r=sqrt((45-4)^2+(42-5)^2)

=sqrt(41^2+37^2)=sqrt3050

r^2=3050

The equation of the circle is

(x-45)^2+(y-42)^2=3050

graph{((x-4)^2+(y-5)^2-1/16)((x-8)^2+(y-1)^2-1/16)(y-x+3)(y-7/9x-7)(x+y-9)=0 [-8.75, 21.25, -4.1, 10.9]}

graph{((x-45)^2+(y-42)^2-3050)(y-x+3)(y-7/9x-7)=0 [-90, 150, -20, 100]}

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