Question

How do you write the equation of the circle through (0,0), (6,8), and (13,7)?

Answer 1

`(x – 8.42)^2 + (y + 0.0645)^2` = 70.9

Explanation:

The center-radius form of the circle equation is in the format `(x – h)^2 + (y – k)^2 = r^2,` with the center being at the point (h, k) and the radius being "r". This form of the equation is helpful, since you can easily find the center and the radius.

Because the radius must be constant, the distance from any point on the circle to the center is the same.
Using the distance formula for points:
Given the two points (x1, y1) and (x2, y2), the distance between these points is given by the formula:

`d = sqrt((x2 – x1)^2 + (y2 - y1)^2)`

`(0 – h)^2 + (0 – k)^2 = (6 – h)^2 + (8 – k)^2 = (13 – h)^2 + (7 – k)^2 = r^2`

1. `(0 – h)^2 + (0 – k)^2 = r^2 ; h^2+ k^2 = r^2`
2. `(6 – h)^2 + (8 – k)^2 = r^2 ; h^2 - 12h + 36 + k^2 - 16k + 64= r^2`
3. `(13 – h)^2 + (7 – k)^2 = r^2 ; h^2 - 26h + 169 + k^2 - 14k + 49= r^2`
   Subtract 1. from 2 & 3.
4. 12h + 36 - 16k + 64= 0 ; -12h -16k + 100 = 0
5. 26h + 169 - 14k + 49= 0 ; -26h -14k + 218 = 0
   Multiply each by GCD of h (12 and 26) and subtract:
   -312h - 416k + 2600 = 0
   -312h - 168k + 2616 = 0
6. 248k - 16 = 0 ; k = -0.0645
   Multiply each by GCD of k (14 and 16) and subtract:
   -168h - 224k + 1400 = 0
   -416h - 224k + 3488 = 0
7. 248h + 2088 = 0 ; h = 8.42
   Put back into 1. to find r:
   `h^2+ k^2 = r^2 ; (8.42)^2 + (-0.0645)^2 = r^2 ; 70.9 + 0.0042 = r^2 ; 70.9 = r^2` ; 8.42= r

CHECK
#(6 – h)^2 + (8 – k)^2 = r^2 ; (6 – (8.42))^2 + (8 – (-0.0645))^2 = 70.9 (-2.42)^2 + (8.0645)^2 = 70.9 # ; 5.856 + 65.036 = 70.9

Therefore, the Circle Equation has its center at (x,y) = (8.42, -0.0645) with a radius of 8.42
`(x – 8.42)^2 + (y + 0.0645)^2 = 70.9`

Second Check:
`(13 – 8.42)^2 + (7 + 0.0645)^2 = r^2 ; (4.58)^2 + (7.0645)^2 = 70.9 ; 20.98 + 49.91 = 70.9`