In order to find the point of intersection, we need to find the equations of the two circles.
Once we know the center [tex] (h,k) [/tex] and radius [tex] r [/tex] of a circle, we can write its expression as
[tex] (x-h)^2+(y-k)^2 = r^2 [/tex]
We know that circle c has center [tex] (0,0) [/tex] and radius 8, so we immediately deduce its equation:
[tex] x^2+y^2 = 64 [/tex]
As for circle k, we know that the segment with endpoints the origin and [tex] (0,18) [/tex] is a diameter. So, the center is the midpoint of this segment, i.e. [tex] (0,9) [/tex]. Moreover, the segment with endpoints the origin and the center is a radius, which is 9 units long.
So, the equation for circle k is
[tex] x^2+(y-9)^2 = 81 [/tex]
So, the intersection points are given by the system
[tex] \begin{cases} x^2+y^2=64 \\ x^2 + (y-9)^2 = 81\end{cases} [/tex]
If we subtract the first equation from the second, we have
[tex] (y-9)^2 - y^2 = 81-64 [/tex]
Which expands to
[tex] y^2-18y + 81 - y^2 = 81-64 \implies -18y = -64 \implies y = \cfrac{32}{9} [/tex]
Which yields the following values for x:
[tex] x^2+ \left(\cfrac{32}{9}\right)^2 = 64 \implies x^2 = 64-\left(\cfrac{32}{9}\right)^2 \implies x = \pm\cfrac{8\sqrt{65}}{9} [/tex]
We are interested in the point belonging to the first quadrant, so we only accept the positive solution. So, we have
[tex] P = \left( \cfrac{8\sqrt{65}}{9}, \cfrac{32}{9} \right) [/tex]
To convert this point in polar coordinates, use the definitions
[tex] r = \sqrt{x^2+y^2},\quad \theta = \arctan\left(\cfrac{y}{x}\right) [/tex]
We already know that [tex] \sqrt{x^2+y^2} = \sqrt{64} = 8 [/tex], because the point lies on circle c.
And finally, we have
[tex] \theta = \arctan\left(\cfrac{32}{9}\cdot\cfrac{9}{8\sqrt{65}}\right) = \arctan\left(\cfrac{4}{\sqrt{65}}\right) \approx 26.39^\circ [/tex]
