# equation of a circle with...

• Feb 21st 2008, 04:45 PM
enji333
equation 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 PM
topsquark
Quote:

Originally Posted by enji333
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
$(x - h)^2 + (y - k)^2 = r^2$

So putting the two points in gives
$h^2 + k^2 = r^2$
and
$(6 - h)^2 + k^2 = r^2$

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:
$(X - h)^2 + (-1 - k)^2 = r^2$

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
$2(X - h) + 2(y - k) \frac{dy}{dx} = 0$

The derivative at (X, -1) is 0, so this becomes
$2(X - h) = 0$

Thus you need to solve the system of equations:
$h^2 + k^2 = r^2$

$(6 - h)^2 + k^2 = r^2$

$(X - h)^2 + (-1 - k)^2 = r^2$

$2(X - h) = 0$

-Dan
• Feb 21st 2008, 04:58 PM
enji333
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 ?

• Feb 21st 2008, 05:07 PM
topsquark
Quote:

Originally Posted by enji333
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 ?

I never gave you a solution. I gave you four equations in the unknowns X, h, k, and r. I'm leaving it for you to solve them. If you have difficulties, please post what you've been able to come up with, and either I or someone else will help you from there.

-Dan
• Feb 21st 2008, 10:57 PM
earboth
Quote:

Originally Posted by enji333
The line y = -1 is a tangent to a circle which passes through (0, 0) and (6, 0).

Find the equation of the circle.

Draw a rough sketch. You can proof by rules of symmetry that the tangent point must be T(3, -1).

Use the equation of the circle. Plug in the coordinates of the 3 points and solve the system of equations for h, k, r.

For confirmation only: (h, k, r) = (3, 4, 5)