When does this equation in $x, y$ and $\theta$ represent a circle?

When does this equation in $x, y$ and $\theta$ represent a circle?

Question

[Choose the one correct answer and explain your reasoning.]

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

---

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

---

1. $0<\theta<\dfrac{\pi}{2}$;

---

1. all values for $\theta$.

Suggestion

What is the general form for the equation of a circle? Can we manipulate this equation into that form?

Solution

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?

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/UQrH6fY9/width/500/height/450/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

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 $\sin \theta >\dfrac{1}{\sqrt2}$.

Hence the answer is (b).