Question

Write the general equation for the circle that passes through the points (-1,2)(4,2)(-3,4)

Answer 1

Answer:

x² + y² - 3x - 13y + 18 = 0

Step-by-step explanation:

Recall that the general equation of a circle looks something like this:

x² + y² + Ax + By + C = 0

substituting each of the points into the equation we get:

for (-1,2)

(-1)² + (2)² + A(-1) + B(2) + C = 0

1 + 4 -A+2B + C = 0

-A + 2B + C + 5 = 0 ------------ eq 1

for (4,2)

(4)² + (2)² + A(4) + B(2) + C = 0

16 + 4 + 4A + 2B + C = 0

4A + 2B + C + 20 = 0 ------------- eq 2

for (-3,4)

(-3)² + (4)² + A(-3) + B(4) + C = 0

9 + 16 -3A + 4B + C = 0

-3A + 4B + C + 25= 0 ----- eq 3

Now we have a system of equations with 3 equations and 3 unknowns.

Solving for A, B and C, we eventually get:

A = -3, B = -13, C = 18

Substituting these into the general equation:

x² + y² + Ax + By + C = 0

x² + y² - 3x - 13y + 18 = 0