Geometry (Coordinates) Help!?

Question

Find the equation of the circle that has center (p,q) and is tangent to the x-axis. Please explain if you can how you got to your solution.

Answer 1

If it's tangent to the x-axis, then the point (p, 0) must satisfy the equation; that is, the circle must have a radius of q.

So it's just:

(x - p)² + (y - q)² = q²

To verify, plug in (p, 0):

(p - p)² + (0 - q)² = q²

0² + q² = q²

If you wanted to write the circle in the "standard form" of Ax² + Bx + Cy² + Dy + E = 0, then it would be:

x² - 2px + p² + y² - 2qy + q² = q²

x² - 2px + y² - 2qy + p² = 0

Answer 2

The standard form of the equation for a circle is

(x - p)^2 + (y - q)^2 = r^2

since the circle is tangent to the x-axis, that means the center is r units away from that axis (ile. it touches in only one place)

so your solution is
(x - p)^2 + (y - q)^2 =q^2

I hope it wasn't a homework problem!

Answer 3

Circle centre (p,q) and radius r has equation:-
(x - p)² + (y - q)² = r²
But in this case r = q and thus circle is:-
(x - p)² + (y - q)² = q²