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

1 Answer
Jun 28, 2016

2x^2 + 2y^2 + 100x + 159y - 1643 = 0

Explanation:

General equation of circle can be represented as x^2+y^2-2hx-2ky+c=0 -> Equation 1

where (h,k) represents the center of the circle , r - radius of the circle, c = h^2 + k^2 - r^2

As we know the given points (4,7) and (7,5) lie on the circle, they must satisfy Equation 1.

Putting the given points (4,7) and (7,5) in the above equation of circle (1) , we get :

4^2+7^2-2h*4-2k*7+c=0

=> -8h - 14k + c = -65 =>

8h + 14k - c = 65 -> Equation 2

7^2+5^2-2h*7-2k*5+c=0

=>-14h - 10k +c=-74 =>

14h + 10k - c=74 -> Equation 3

Since the centre of the circle is (h,k) lies on the line y = 7/4x + 4, (h,k) must satisfy the equation of line.

k = 7/4*(h) + 4 =>

7h - 4k = -16 -> Equation 4

Subtract Equation 2 and 3

=> (14h - 8h) + (10k - 14k) + (-c+c) = 74 - 65#

6h -4k = 9 -> Equation 5

Add Equation 4 and 5

=>(-7h+6h) + (4k -4k) = (16+9) -> Equation 5

-h = 25 => h= -25

-4k= 9 - (6*(-25)

k = -159/4

Solving for c in Equation 3, we get c = 14*(-25) + 10*(-159/4) - 74

c= -3286/4 = -1643/2

Circle equation is:

x^2+y^2-2(-25)x-2(-159/4)y- 1643/2=0

x^2+y^2+50x-(-159/2)y - 1643/2=0

2x^2 + 2y^2 + 100x + 159y - 1643 = 0