Math question?

Question

Determine the equation of the circle passing through (3,0), (-3,0), and (0,9), using the general equation x² + y² +ax+by+c = 0.

Answer 1

replace x and y of each point into the equation then you will get 3 equations and three unknowns

(3,0)
9 + 0 + 3a + 0 + c = 0
3a + c = -9 ...... (1)

(-3,0)
9 + 0 - 3a + 0 + c = 0
-3a + c = -9 ......(2)

(0, 9)
0 + 81 + 0 + 9b + c =0
9b + c = -81 .....(3)

solving
6a = 0 => a = 0
replacing a in (1) we get c=-9

(3)=>9b - 9 = -81 => b = -8

therefore the equation will be:

x^2 + y^2 -8y -9 = 0

Answer 2

The previous answer is a nice one using pure algebra. Here is another approach relying more on geometric intuition and less algebra. Too bad I can't show a picture.

The circle passing through all 3 points has its center on the y-axis between (0,0) and (0,9). Let the radius be r and the center coordinates are therefore (0,9-r). (Can you visualize that circle which is up a little on the Y-axis?) Equation for the circle is therefore:

X^2 + (Y-(9-r))^2 = r^2
X^2 + (Y-9+r)^2 = r^2 .

We now take the coordinates (3,0), substitute into the equation and solver for r:

9 + (0-9+r)^2 = r^2,
9 + (81-18r+r^2) = r^2,
90 - 18r + r^2) = r^2,
90 = 18r,
r=5.

So the equation is:

X^2 + (Y-(9-5))^2 = 5^2, or
X^2 + (Y-4)^2 = 25.

If you expand it out, it will yield the same answer as the previous answer:

X^2 + Y^2 - 8Y -9 = 0.

Answer 3

Maybe I'm just confused but isn't a circle supposed to have a an equal radius from it's center point?? This one seems to have a radius of 3 (assuming the center is 0,0) from (3,0) to (-3,0) but I don't understand how the 0,9 comes into play...please advise

Answer 4

Same as kate's answer