1. ## polar coordinates

A circle C has center at the origin and radius 3. Another circle K has a diameter with one end at the origin and the other end at the point (0, 13). The circles C and K intersect in two points. Let P be the point of intersection of C and K which lies in the first quadrant. Let $(r,\theta)$ be the polar coordinates of P, chosen so that $r$is positive and $0 \leq \theta \leq \pi/2$

Find $r$ and $\theta$

2. Originally Posted by viet
A circle C has center at the origin and radius 3. Another circle K has a diameter with one end at the origin and the other end at the point (0, 13). The circles C and K intersect in two points. Let P be the point of intersection of C and K which lies in the first quadrant. Let $(r,\theta)$ be the polar coordinates of P, chosen so that $r$is positive and $0 \leq \theta \leq \pi/2$

Find $r$ and $\theta$
So you are looking for the intersections of the circles
$x^2 + y^2 = 9$
and
$x^2 + \left (y - \frac{13}{2} \right )^2 = \left ( \frac{13}{2} \right ) ^2$

From the first equation $x^2 = 9 - y^2$. Inserting this into the second equation gives:
$(9 - y^2) + \left (y - \frac{13}{2} \right ) ^2 = \frac{169}{4}$

Solve this for y. I get:
$y = \frac{9}{13}$

There are two x values for this y value:
$x = \pm \sqrt{9 - \left ( \frac{9}{13} \right ) ^2} = \pm \frac{12}{13} \sqrt{10}$

Thus the point of intersection in QI is
$\left ( \frac{12}{13} \sqrt{10}, \frac{9}{13} \right )$

I leave putting this into polar coordinates to you.

-Dan

3. Hello, viet!

A circle $C$ has center at the origin and radius 3.
Another circle $K$ has a diameter with one end at the origin and the other end at the point (0, 13).
The circles $C$ and $K$ intersect in two points.

Let $P$ be the point of intersection of $C$ and $K$ which lies in the first quadrant.
Let $(r,\,\theta)$ be the polar coordinates of $P$, chosen so that $r$ is positive and $0 \leq \theta \leq \pi/2$

Find $r$ and $\theta$
If you are familiar with polar equations of circle, it's easier.

Circle $C$ has the equation: . $r\:=\:3$

Circle $K$ has the equation: . $r \:=\:13\sin\theta$

They intersect where: . $13\sin\theta \:=\:3$

Hence, we have: . $\sin\theta \:=\:\frac{3}{13}\quad\Rightarrow\quad\theta \:=\:\sin^{-1}\left(\frac{3}{13}\right) \:=\:0.232868178 \:\approx\:0.233$

Therefore, point $P$ is: . $3,\,0.233)$