JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

Two diameters of the circle

3{x^2} + 3{y^2} - 6x - 18y - 7 = 0

are along the lines

3x + y = {c_1}

and

x - 3y = {c_2}

. Then, the value of

{c_1}{c_2}

View answer and explanation

Correct option: A

- 48

Option A (

- 48

) is correct because it is the product of the constants

c_1

and

c_2

obtained by substituting the center of the circle into the diameter equations.

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