Question

A circle c has center at the origin and radius 8. another circle k has a diameter with one end at the origin and the other end at the point (0,18).(0,18). 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)(r,θ) be the polar coordinates of p, chosen so that rr is positive and 0≤θ≤2.0≤θ≤2. find rr and θθ.

Answer 1

In order to find the point of intersection, we need to find the equations of the two circles.

Once we know the center $(h,k)$ and radius $r$ of a circle, we can write its expression as

$(x-h)^2+(y-k)^2 = r^2$

We know that circle c has center $(0,0)$ and radius 8, so we immediately deduce its equation:

$x^2+y^2 = 64$

As for circle k, we know that the segment with endpoints the origin and $(0,18)$ is a diameter. So, the center is the midpoint of this segment, i.e. $(0,9)$. 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

$x^2+(y-9)^2 = 81$

So, the intersection points are given by the system

$\begin{cases} x^2+y^2=64 \\ x^2 + (y-9)^2 = 81\end{cases}$

If we subtract the first equation from the second, we have

$(y-9)^2 - y^2 = 81-64$

Which expands to

$y^2-18y + 81 - y^2 = 81-64 \implies -18y = -64 \implies y = \cfrac{32}{9}$

Which yields the following values for x:

$x^2+ \left(\cfrac{32}{9}\right)^2 = 64 \implies x^2 = 64-\left(\cfrac{32}{9}\right)^2 \implies x = \pm\cfrac{8\sqrt{65}}{9}$

We are interested in the point belonging to the first quadrant, so we only accept the positive solution. So, we have

$P = \left( \cfrac{8\sqrt{65}}{9}, \cfrac{32}{9} \right)$

To convert this point in polar coordinates, use the definitions

$r = \sqrt{x^2+y^2},\quad \theta = \arctan\left(\cfrac{y}{x}\right)$

We already know that $\sqrt{x^2+y^2} = \sqrt{64} = 8$, because the point lies on circle c.

And finally, we have

$\theta = \arctan\left(\cfrac{32}{9}\cdot\cfrac{9}{8\sqrt{65}}\right) = \arctan\left(\cfrac{4}{\sqrt{65}}\right) \approx 26.39^\circ$