# Circle analytic geometry help?

• Sep 18th 2008, 06:11 PM
realintegerz
Circle analytic geometry help?
1) Find an eq. for a circle that has its center on the y-axis, that has a radius of 4, and that has only one point on it whose x-coordinate us the same as it's y-coordinate

It would be great if you could help on this one, one of my questions is...since its centered on the y-axis and its radius is 4, as i go up and down the y-axis, wouldn't there be 2 points that have the same value for x and y? (that is including the negative)

What i mean is..for example lets say its centered on (0,3).. wouldnt (-4,3) & (4,3) be on the circle ? maybe im thinking wrong

2) 2 circles are symmetric across the y-axis. The points (7,5) and (3,1) are on endpoints of a diameter of one of the circles. What are the eq.'s of the circles?

Does symmetric mean that they are sticking to each other? I can find the eq. of one of the circles from the endpoints given, but what does them begin symmetric mean, i think a hint here would lead me to the answer

3) A circle with a center at (4,5) is 1st reflected across the y-axis and then across the x-axis. The point (-2,-6) is on this new circle. What are the eq.'s of each of the circles and what is the eq. of the line through the centers of the circles?

I'm lost on this one....
• Sep 18th 2008, 07:06 PM
icemanfan
1. The question is not whether you have two points (x1, y1) and (x2, y2) with the property that $|x_1| = |x_2|$ and $|y_1| = |y_2|$, but whether there is exactly one point on the circle (x, y) such that x = y. This means that the circle is tangent to the line y = x. Also, this point must be 4 units away from the center, which is at (0, k) for some value of k, since it is on the y-axis.

2. If two circles are symmetric across an axis, this means that you can obtain one of the circles by flipping the other over that axis. It also means that the centers of both circles are the same distance away from the axis of symmetry. This information should help you find the second circle.

3. Work backwards from the new circle. First flip it back over the x-axis. If (-2, -6) is on the new circle, then the point (-2, 6) is on the circle flipped back over the x-axis. Now flip it back over the y-axis. Since (-2, 6) was on the previous circle, (2, 6) must be on the original circle. The key is that when you flip an equation over the y-axis, points (x, y) are translated to (-x, y), and when you flip an equation over the x-axis, points (x, y) are translated to (x, -y).
• Sep 18th 2008, 07:39 PM
Soroban
Hello, realintegerz!

Quote:

1) Find an eq. for a circle that has its center on the y-axis, that has a radius of 4,
and that has only one point on it whose x-coordinate is the same as its y-coordinate

Code:

                |               * * *           *    |    *         *      |      *       *        |        *                 |       *        |        *       *    (0,k)*        *       *        |        *                 |       *        |        *         *      |      *           *    |    o P               * * *                 |                 |   - - - - - - - + - - - - - - -                 |

The circle has center (0,k) and radius 4.

It equation is: . $x^2 + (y-k)^2 \:=\:16$ .[1]

There is one point $P$ where the x-coordinate and y-coordinate are equal.
. . $\text{That is: }\:x = y.$

Substitute into [1]: . $x^2 + (x-k)^2 \:=\:16 \quad\Rightarrow\quad 2x^2 - 2kx + k^2-16 \:=\:0$

The Quadratic Formula gives us: . $x \;=\;\frac{k \pm\sqrt{32-k^2}}{2}$

It would seem that there are two such points.

Since we are told there is only one such point, the discriminant must be zero.
. . That is: . $32 - k^2 \:=\:0$

Hence: . $k^2 \:=\:32\quad\Rightarrow\quad k \:=\:\pm4\sqrt{2}$

There are two possible circles:

. . $\begin{array}{cc}\text{One above the x-axis:} & x^2 + (y-4\sqrt{2})^2 \:=\:16 \\\text{One below the x-axis:} & x^2 + (y+4\sqrt{2})^2 \:=\:16\end{array}$

• Sep 18th 2008, 08:13 PM
realintegerz
so right now im just having problems with #3

can anyone help on that?

i got the equations for the circles

(x+4)^2 + (y+5)^2 = 37
(x-4)^2 + (y-5)^2 = 37

but forgot how to get the line equation
• Sep 18th 2008, 08:39 PM
realintegerz
never mind, i got it