A variable circle $C$ has the equation $x^2+y^2-2\left(t^2-3 t+1\right) x-2\left(t^2+2 t\right) y+t=0$, where $t$ is $a$ parameter. If the power of point $(a, b)$ w.r.t. the circle $C$ is constant, then the ordered pair $(a, b)$ is

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

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Analyzing +4 | -1

A variable circle

C

has the equation

x^2+y^2-2\left(t^2-3 t+1\right) x-2\left(t^2+2 t\right) y+t=0

, where

t

a

parameter. If the power of point

(a, b)

w.r.t. the circle

C

is constant, then the ordered pair

(a, b)

View answer and explanation

Correct option: C

\left(-\frac{1}{10}, \frac{1}{10}\right)

Option C

\left(-\frac{{1}}{{10}}, \frac{{1}}{{10}}\right)

is correct because the power of the point

(a, b)

with respect to the circle is constant only when

a + b = 0

and

6a - 4b + 1 = 0

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