This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

## Equation of a circle- ISI MStat Year 2016 PSA Question 9

Given \( \theta \) in the range \( 0 \leq \theta<\pi,\) the equation \( 2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0\) represents a circle for all \( \theta\) in the interval

- \( 0 < \theta <\frac{\pi}{3} \)
- \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)
- \( 0 < \theta <\frac{\pi}{2} \)
- \( 0 \le \theta <\frac{\pi}{2} \)

**Key Concepts**

Equation of a circle

Trigonometry

Basic Inequality

## Check the Answer

But try the problem first…

Answer: is \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

## Try with Hints

First hint

Complete the Square.

Second Hint

We get ,

\(2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3)) \)

\(6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2} \)

Final Step

We are given that \( 0 \leq \theta<\pi,\) . So, \( {\sin^2 \theta} \geq \frac{1}{2} \) \( \Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4} \).

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