The centre of the circ le $x = 2 + 3\cos {\rm{\theta }},y = 3\sin \theta - 1$ is

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

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Understanding +4 | -1

The centre of the circ le

x = 2 + 3\cos \mathrm{\theta },y = 3\sin \theta - 1

View answer and explanation

Correct option: B

\left( {2, - 1} \right)

The correct answer is B (

\left( {2, - 1} \right)

) because the parametric equations directly provide the center of the circle in the form

x = h + r\cos{\theta}

y = k + r\sin{\theta}

, where

(h, k)

is the center.

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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