Let $S$ be the circle in the $x y$-plane defined by the equation $x^2+y^2=4$. Let $P$ be a point on the circle with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at $M$ and $N$. Then the mid-point of the line segment MN must lie on the curve

$x^{2 / 3}+y^{2 / 3}=2^{4 / 3}$

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

Let

S

be the circle in the

x y

-plane defined by the equation

x^2+y^2=4

. Let

P

be a point on the circle with both coordinates being positive. Let the tangent to

S

P

intersect the coordinate axes at

M

and

N

. Then the mid-point of the line segment MN must lie on the curve

View answer and explanation

Correct option: D

x^2+y^2=x^2 y^2

Option D (

x+2 \sqrt{2} y=6

) is correct because it is a common tangent to both 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