If the area of the circle $4{x^2} + 4{y^2} - 8x + 16y + k = 0$ is $9\pi $ sq unit, then the value of $k$ is

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

If the area of the circle

4{x^2} + 4{y^2} - 8x + 16y + k = 0

9\pi

sq unit, then the value of

k

View answer and explanation

Correct option: C

- 16

Option C (-16) is correct because it is the value of k that makes the area of the given circle equal to 9π.

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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