Given $2$ circles and radius, find the angle.

Question

A circle $C$ has center at the origin and radius $6$. Another circle $K$ has a diameter with one end at the origin and the other end at the point $(0, 10)$. 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 $(r, θ)$ be the polar coordinates of $P$, chosen so that $r$ is positive and $0\le θ\le 2$. Find $r$ and $θ$.

I already figured out that r is 6, but I have no clue how to find the angle. Could someone help?

Answer 1

Draw a sketch. (Recognition is by visual input).

$$ x^2+y^2=36$$ Due to property of circles that product of segments is constant $$ y (1 0-y)= x^2 $$ Solve to find $$ y=\frac{36}{10}=3.6$$ Can you now find $x$?

Answer 2

Equation for circle $A$ is easy: $x^2+y^2=6^2$. Because both circles crossed at $P$, $r=6$, as you guessed. Equation for circle $K$ is: $(x-5)^2+(y-0)^2=5^2$ or $x^2+y^2=10x$, because by given date, it can be determined that its radius is $5$ and the coordinate of radius is $(5,0)$.

Suppose the coordinates for P is $(x_1,y_1)$. Since it's an intersection,

$x_1^2+y_1^2=6^2$ and $x_1^2+y_1^2=10x_1$, solution of which gave: $x_1=\frac{18}{5}$ and $x_1=\frac{24}{5}$.

Thus, by polar coordinates, $r\sin{\theta}=y_1$ and $r\cos{\theta}=x_1$, therefore, $\tan{\theta}=\frac{y_1}{x_1}=\frac{24}{18}=\frac{4}{3}$.