A circle has a center that falls on the line y = 7/9x +7 and passes through ( 4 ,5 ) and (3 ,7 ). What is the equation of the circle?

1 Answer
Oct 11, 2016

(sqrt(22570)/10)^2 = (x - -99/10)^2 + (y - -7/10)^2

Explanation:

The standard form for the equation of a circle is:

r^2 = (x - h)^2 + (y - k)^2

where, (x,y) is any point on the circle, (h, k) is the center point, and r is the radius.

Using the two points we can write two equations:

r^2 = (4 - h)^2 + (5 - k)^2
r^2 = (3 - h)^2 + (7 - k)^2

Because r^2 = r^2, the right sides of these equations are equal:

(4 - h)^2 + (5 - k)^2= (3 - h)^2 + (7 - k)^2

I will expand the squares using the pattern (a - b)^2 = a^2 - 2ab + b^2:

16 - 8h + h^2 + 25 - 10k + k^2= 9 - 6h + h^2 + 49 - 14k + k^2

Combine like terms:

4k= 2h + 17

We substitute the point (h, k) into the given equation:

k = 7/9h + 7

Solve these two equation for h then k:

4(7/9h + 7)= 2h + 17

28/9h + 28 = 2h + 17

10/9h = -11

h = -99/10

4k= 2(-99/10) + 17

4k= -198/10 + 170/10

4k= -28/10

k = -7/10

Check:

-7/10 = 7/9(-99/10) + 7

-7/10 = -77/10 + 70/10

This checks

Use one of the equations of the circle and the center (-99/10, -7/10) to find r:

r^2 = (4 - -99/10)^2 + (5 - -7/10)^2

r^2 = (40/10 - -99/10)^2 + (50/10 - -7/10)^2

r^2 = 19321/100 + 3249/100

r^2 = 22570/100

r = sqrt(22570)/10

The equation of the circle is:

(sqrt(22570)/10)^2 = (x - -99/10)^2 + (y - -7/10)^2