JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

If the equation

\lambda {x^2} + \left( {2\lambda - 3} \right){y^2} - 4x - 1 = 0

represents a circle, then its radius is

View answer and explanation

Correct option: C

\frac{{\sqrt 7 }}{3}

Option C (

\frac{{\sqrt 7 }}{3}

) is correct because after substituting

\lambda = 3

into the given equation and simplifying, the radius of the resulting circle is calculated to be

\frac{{\sqrt 7 }}{3}

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