Find the equation of a circle with the centre on the x-axis. The circles passes through points (2,2) & (-4,-4)?
I just can't seem to solve this problem, so far i've done this:
Find the equation of a circle with the centre on the x-axis. The circles passes through points (2,2) & (-4,-4)
Since it's centre is on the x axis, the y coordinate of the center is 0.
Let's name the points A and B:
A(2, 2)
B(-4, -4)
So for A:
q=0 therefore,
(2-p)^2+(2-0)^2 = r^2
4 - 4p +p^2 + 4 = r^2
p^2 - 4p + 8 = r^2
And for B:
(-4-p)^2+(-4-0)^2 = r^2
16 + 8p + p^2 +16 = r^2
p^2 + 8p + 32 = r^2
Now since both of these points can be found on the circle, we know that the radii are the same, so we can equal them.
p^2 - 4p + 8 = p^2 + 8p + 32
p^2 + 8p + 32 - p^2 + 4p - 8 = 0
12p = -24
p = -2
So
p = -2
q = 0
Therefore the equation of the circle is
(x + 2)^2 + y^2 = r^2
Now the thing i don't get is: how do you get the radius?