Q 69 - Calculus

Question

Modify the proof of Theorem $9.8$ to show that the graph of the equation $r=k\sin \theta +l\cos \theta$ is a circle. Find the center and radius in terms of $k$  and $l$ . 

Step-by-Step Solution

Answer

The equation $r=k\sin \left(\theta \right)+l\cos \left(\theta \right)$ is converted in rectangular coordinates as following :-

${\left(x-\frac{l}{2}\right)}^2+{\left(y-\frac{k}{2}\right)}^2=\frac{k^2+l^2}{4}$

Clearly, This is the equation of circle.

By comparing this with the general equation of the circle ${\left(x-h\right)}^2+{(y-k)}^2=r^2$, where $\left(h,k\right)$ is center and $r$ is radius of circle, we have :-

$\left(\frac{l}{2},\frac{k}{2}\right)$ is center and $\frac{\sqrt{k^2+l^2}}{2}$ is radius of the circle.

Step 1. Given Information

We have given the following equation :-

$r=k\sin \theta +l\cos \theta$.

We have to prove that this is the equation of circle. Also we have to find center and radius of the circle in terms of $k$ and $l$.

Step 2. To prove the given equation is equation of a circle

We have given the following equation :-

$r=k\sin \theta +l\cos \theta$.

We know that :-

$\cos \theta =\frac{x}{r}$ and $sin\theta =\frac{y}{r}$.

By putting these values we have :-

$$\begin{gathered} r=k\frac{y}{r}+l\frac{x}{r} \\ \Rightarrow r=\frac{lx+ky}{r} \\ \Rightarrow r^2=lx+ky \end{gathered}$$

Also we know that :-

$r^2=x^2+y^2$

That is :-

$x^2+y^2=lx+ky$

We can completing the squares as following :-

$$\begin{gathered} x^2+y^2-lx-ky=0 \\ \Rightarrow \left(x^2-lx+\frac{l^2}{4}\right)+\left(y^2-ky+\frac{k^2}{4}\right)=\frac{l^2}{4}+\frac{k^2}{4} \\ \Rightarrow {\left(x-\frac{l}{2}\right)}^2+{\left(y-\frac{k}{2}\right)}^2=\frac{k^2+l^2}{4} \end{gathered}$$

Now we can see that this is the equation of circle.

So we can conclude that the given equation is the equation of circle.

Step 3. To find center and radius

We converted the given equation as following :-

${\left(x-\frac{l}{2}\right)}^2+{\left(y-\frac{k}{2}\right)}^2=\frac{k^2+l^2}{4}$

By comparing this equation with the general equation of circle ${\left(x-h\right)}^2+{(y-k)}^2=r^2$, where $\left(h,k\right)$ is the center of the circle and $r$ is radius.

Then we have :-

$\left(\frac{l}{2},\frac{k}{2}\right)$ is the center of the circle and $\frac{\sqrt{k^2+l^2}}{2}$is radius.