JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

a > 2b > 0

, then the positive value of

m

for which

y = mx - b\sqrt {1 + {m^2}}

is a common tangent to

{x^2} + {y^2} = {b^2}

and

{\left( {x - a} \right)^2} + {y^2} = {b^2}

, is

View answer and explanation

Correct option: A

\frac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}

Option A (

\frac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}

) is correct because it is derived from the tangency condition of a line to a circle.

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