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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

Check the Answer


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}.

ISI MStat 2016 PSA Problem 9
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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

Check the Answer


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}.

ISI MStat 2016 PSA Problem 9
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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