JEE MATHEMATICS MCQ: CIRCLES

1 Medium

1 of 1

Analyzing +4 | -1

If a line,

y = mx + c

is a tangent to the circle,

{(x - 3)^2} + {y^2} = 1

and it is perpendicular to a line

{L_1},

where

{L_1}

is the tangent to the circle,

{x^2} + {y^2} = 1

at the point

\left( {\frac{1}{{\sqrt 2 }},\,\frac{1}{{\sqrt 2 }}} \right)

; then

View answer and explanation

Correct option: D

{c^2} + 6c + 7 = 0

Option D (

{c^2} + 6c + 7 = 0

) is correct because it represents the equation derived from the tangency condition of the line to the given circle.

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