JEE MATHEMATICS MCQ: CIRCLES

1 Hard

1 of 1

Analyzing +4 | -1

Two concentric circles of which smallest is

x^2+y^2=4

, have the difference in radii as

d

, if line

y=x+1

cuts the circles in real points, then

d

lies in the interval

View answer and explanation

Correct option: A

\left(-\infty,-2-\frac{1}{\sqrt{2}}\right) \cup\left(-2+\frac{1}{\sqrt{2}}, \infty\right)

The correct answer is A

\left(-\infty,-2-\frac{{1}}{{\sqrt{{2}}}}}\right) \cup\left(-2+\frac{{1}}{{\sqrt{{2}}}}, \infty\right)

because it represents the interval for the difference in radii such that the line intersects both circles in real points.

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