Chapter 8.2, Problem 57E

To determine

To show that the graph of below polar equation is a circle and to find it’s centre and radius -

  $r=a\cos \theta +b\sin \theta$

Answer to Problem 57E

The graph of polar equation $r=a\cos \theta +b\sin \theta$ is a circle with Centre as $\left(\frac{a}{2},\frac{b}{2}\right)$ and radius as $r=\frac{1}{2}\sqrt{a^2+b^2}$ .

Explanation of Solution

Given: Polar equation: $r=a\cos \theta +b\sin \theta$

Formula Used:

A polar equation is any equation that describes a relation between r and θ , where r represents the distance from the pole (origin) to a point on a curve, and θ represents the counter-clockwise angle made by a point on a curve, the pole, and the positive x-axis.

Also, $x=r\cos \theta$ and $y=r\sin \theta$

  $r=\sqrt{x^2+y^2}$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$

Equation of Circle is given as:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ ; where Centre is $\left(h,k\right)$ and radius is $r$

Calculation:

Given Polar equation is $r=a\cos \theta +b\sin \theta$

Converting the above polar equation to rectangular form using the above relationships, we have:

  $r=a\cos \theta +b\sin \theta$

  $\begin{array}{l}r=a\left(\frac{x}{r}\right)+b\left(\frac{y}{r}\right)\\ r^2=ax+by\\ x^2+y^2=ax+by\\ x^2-ax+y^2-by=0\end{array}$

Now, completing the squares of the above equation, we have:

  $\begin{array}{l}\left(x^2-ax+\frac{a^2}{4}\right)-\frac{a^2}{4}+\left(y^2-by+\frac{b^2}{4}\right)-\frac{b^2}{4}=0\\ {\left(x-\frac{a}{2}\right)}^2+{\left(y-\frac{b}{2}\right)}^2=\frac{a^2}{4}+\frac{b^2}{4}\end{array}$

The above equation is of the form - ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

So, the Centre is $\left(h,k\right)=\left(\frac{a}{2},\frac{b}{2}\right)$

And radius is $r$ -

  $\begin{array}{l}r=\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\\ r=\frac{1}{2}\sqrt{a^2+b^2}\end{array}$

Thus, proved that the equation $r=a\cos \theta +b\sin \theta$ is a Circle with Centre as $\left(\frac{a}{2},\frac{b}{2}\right)$ and radius as $r=\frac{1}{2}\sqrt{a^2+b^2}$ .

Conclusion:

Hence, proved that the equation $r=a\cos \theta +b\sin \theta$ is a Circle with Centre as $\left(\frac{a}{2},\frac{b}{2}\right)$ and radius as $r=\frac{1}{2}\sqrt{a^2+b^2}$ .