JEE MATHEMATICS MCQ: CIRCLES

1 Hard

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Analyzing +4 | -1

The values of

\theta

\left[ {0,2\pi } \right]

so that circles

{x^2} + {y^2} + 2\left( {\sin \alpha } \right)x + 2\left( {\cos \alpha } \right)y + {\sin ^2}\theta = 0

always lies inside the square of unit side length, is /are

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the correct range for

\theta

\left[ {\pi /3,2\pi /3} \right] \cup \left[ {4\pi /3,5\pi /3} \right]

, which is not listed in the options.

More Questions from CIRCLES

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{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}},

x

\left( {\alpha ,2{\alpha ^2}} \right)

{x^2} + {y^2} = 4,

{x^2} + {y^2} - 2x + 4y - 4 = 0

{x^2} + {y^2} = {a^2}

\left( {{x_1},{y_1}} \right)

\quad \alpha, \beta

\gamma

P, Q

x^2+y^2=1

A

(-1,0)

A P, A Q

A R

G P

\cos \alpha / 2, \cos \beta / 2

\cos \gamma / 2

x+y=6

x+2 y=4

x+y=5

x-y=1

9 \pi