JEE MATHEMATICS MCQ: CIRCLES

1 Easy

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Application +4 | -1

{P_1},{P_2},{P_3}

are the perimeter of the three circles

{x^2} + {y^2} + 8x - 6y = 0,\;4{x^2} + 4{y^2} - 4x - 12y - 186 = 0

and

{x^2} + {y^2} - 6x + 6y - 9 = 0

respectively, then

View answer and explanation

Correct option: B

{P_1} < {P_3} < {P_2}

Option B (

{P_1} < {P_3} < {P_2}

) is correct because it accurately compares the perimeters of the three circles based on their calculated radii.

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