Review question

# When does this equation in $x, y$ and $\theta$ represent a circle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7378

## Solution

Given $\theta$ in the range $0\leq \theta < \pi$, the equation $x^2+y^2+4x \cos\theta+8y \sin\theta+10=0$ represents a circle for

1. $0<\theta<\dfrac{\pi}{3}$;

2. $\dfrac{\pi}{4}<\theta<\dfrac{3\pi}{4}$;

3. $0<\theta<\dfrac{\pi}{2}$;

4. all values for $\theta$.

This applet shows the curve $x^2+y^2+4x \cos\theta+8y \sin\theta+10=0$. We can vary $\theta$ in the range $0\leq \theta < \pi$.

What happens to the centre and radius of the circle?

By completing the square we obtain $(x+2 \cos\theta)^2- 4\cos^2\theta+(y + 4\sin\theta)^2-16\sin^2\theta+10=0.$

Then by rearranging we get $(x+2\cos\theta)^2+(y+4\sin\theta)^2=4\cos^2\theta + 16\sin^2\theta-10.$

Since $\sin^2\theta + \cos^2\theta=1$, we see that the right-hand side becomes $4\cos^2\theta + 16\sin^2\theta-10 = 12\sin^2\theta-6.$

So the equation defines a circle of radius $\sqrt{12 \sin^2\theta-6}$, for $\sin^2\theta>\dfrac{1}{2}$.

In the range $0\leq \theta < \pi$, we have $\sin\theta\geq0$ so we need $\sin \theta >\dfrac{1}{\sqrt2}$.