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 \(\sin \theta >\dfrac{1}{\sqrt2}\).

Graph showing sine x in the region between 0 and pi and the line y = 1 over root 2, and marking the points of intersection at x = pi over 4 and x= 3 pi over 4

Hence the answer is (b).