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

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- Feb 21st 2008, 04:45 PMenji333equation of a circle with...
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 - Feb 21st 2008, 04:53 PMtopsquark
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 - Feb 21st 2008, 04:58 PMenji333
Hi,

First of all, sorry for double posting (Crying)

About your solution. I'm not sure about the final answer you gave. Would you be able to show me the final answer in the form (x - a)^2 + (y - b)^2 = r^2 ?

Appreciate help again, thanks for your answer also - Feb 21st 2008, 05:07 PMtopsquark
- Feb 21st 2008, 10:57 PMearboth