JEE MATHEMATICS MCQ: CIRCLES

1 Medium

1 of 1

Application +4 | -1

If the polar of a point

(\alpha, \beta)

with respect to any one of the circle

x^2+y^2-2 k x+3=0

, where

k

is a variable, always passes through a fixed point, whatever be the value of

k

, then the fixed point is

View answer and explanation

Correct option: A

\left(-\alpha, \frac{1}{\beta}\left(\alpha^2-3\right)\right)

Option A (

\left(-\alpha, \frac{{1}}{{\beta}}\left(\alpha^2-3\right)\right)

) is correct because the polar equation must be independent of k, leading to two equations that can be solved for the fixed point.

More Questions from CIRCLES

Practice All Chapter MCQs →

{\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