JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Analyzing +4 | -1

If the chords of contact of the tangents from a point on the circle

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

to the circle

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

touch the circle

{x^2} + {y^2} = {c^2},

then the roots of the equation

a{x^2} + 2bx + c = 0

, are

View answer and explanation

Correct option: B

Real and equal

The roots are real and equal because the condition for the chord of contact to touch the inner circle leads to

b^2 = ac

, which implies the discriminant of the quadratic equation is zero.

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