Q 68 - Calculus

Question

Finish the proof of Theorem $9.8$ by showing that the graph of the equation $r=2k\sin \theta$ is a circle with center $\left(0,k\right)$ and radius $k$ tangent to the x-axis for any $k\ne 0$. 

Step-by-Step Solution

Answer

The equation $r=2k\sin \theta$ converted in the Cartesian coordinates as following :-

$x^2+{(y-k)}^2=k^2$.

By comparing this with the general equation of the circle ${\left(x-h\right)}^2+{(y-k)}^2=r^2$, we get that the graph of given equation $r=2k\sin \theta$ is a circle with center $\left(0,k\right)$ and radius $k$.

Step 1. Given Information

We have given the following equation :-

$r=2k\sin \theta$

We have to prove that this is the equation of circle with center $\left(0,k\right)$ and radius $k$.

Step 2. Required proof

The given equation is :-

$r=2k\sin \theta$.

This equation is in polar coordinates.

In the theorem $9.8$, we write this equation in rectangular coordinates as following :-

$x^2+{(y-a)}^2=a^2$.

We know that the general equation of circle is :-

${\left(x-h\right)}^2+{(y-k)}^2=r^2$, where $\left(h,k\right)$ is the center of circle of radius $r$.

By comparing both of these equations, we can easily get that that :-

$\left(h,k\right)=\left(0,k\right)$ and $r=k$.

Hence we can conclude that the given equation is the equation of circle with center $\left(0,k\right)$ and radius $k$, tangent to the x-axis for any $k\ne 0$.