Cheenta
Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More

# ISI MStat 2016 PSA Problem 9 | Equation of a circle

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

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

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

## Try with Hints

Complete the Square.

We get ,

$$2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3))$$
$$6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2}$$

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

## Subscribe to Cheenta at Youtube

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

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

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

## Try with Hints

Complete the Square.

We get ,

$$2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3))$$
$$6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2}$$

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

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.