Chapter H.1, Problem 57E

To determine

To show:

The polar equation $r=a\sin \theta +b\cos \theta$ is equation of circle and tell the value of the centre and radius.

Explanation of Solution

Given information:

  $r=a\sin \theta +b\cos \theta$

Proof:

Know that:

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$

Given that:

  $\begin{array}{l}r=a\sin \theta +b\cos \theta \\ ab\ne 0\\ r=a\left(\frac{y}{r}\right)+b\left(\frac{x}{r}\right)\\ r^2=ay+bx\\ x^2+y^2=ay+bx\left[\because r^2=x^2+y^2\right]\\ x^2-bx+y^2-ay=0\\ \left(x^2-bx+\frac{b^2}{4}\right)+\left(y^2-ay+\frac{a^2}{4}\right)=\frac{a^2}{4}+\frac{b^2}{4}\\ {\left(x-\frac{b}{2}\right)}^2+{\left(y-\frac{a}{2}\right)}^2=\frac{a^2}{4}+\frac{b^2}{4}\end{array}$

So, on comparing the simplified equation with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , get that it is polar equation of the circle.

Compared it with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ whose centre and radius are given as:

  $\begin{array}{l}c=\left(h,k\right)\\ r^2=a^2+b^2\end{array}$

So, the centre and radius of ${\left(x-\frac{b}{2}\right)}^2+{\left(y-\frac{a}{2}\right)}^2=\frac{a^2}{4}+\frac{b^2}{4}$ is calculated as:

The centre is: $c=\left(h,k\right)=\left(\frac{a}{2},\frac{b}{2}\right)$

Radius:

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