Hi,
Forgot how to do this.
The line y = -1 is a tangent to a circle which passes through (0, 0) and (6, 0).
Find the equation of the circle.
Any help appreciated
Well the circle equation is
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$
So putting the two points in gives
$\displaystyle h^2 + k^2 = r^2$
and
$\displaystyle (6 - h)^2 + k^2 = r^2$
The circle is tangent to the line y = -1. This tells us a few things. First, the circle has a point (X, -1) on it:
$\displaystyle (X - h)^2 + (-1 - k)^2 = r^2$
We also know that the derivative of the circle equation is 0 at y = -1, since y = -1 is a horizontal line. So taking the derivative of the circle equation we get
$\displaystyle 2(X - h) + 2(y - k) \frac{dy}{dx} = 0$
The derivative at (X, -1) is 0, so this becomes
$\displaystyle 2(X - h) = 0$
Thus you need to solve the system of equations:
$\displaystyle h^2 + k^2 = r^2$
$\displaystyle (6 - h)^2 + k^2 = r^2$
$\displaystyle (X - h)^2 + (-1 - k)^2 = r^2$
$\displaystyle 2(X - h) = 0$
-Dan