If the equation of tangent to the circle ${x^2} + {y^2} - 2x + 6y - 6 = 0$ parallel to $3x - 4y + 7 = 0$ is $3x - 4y + k = 0$, then the value of $k$ are

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

If the equation of tangent to the circle

{x^2} + {y^2} - 2x + 6y - 6 = 0

parallel to

3x - 4y + 7 = 0

3x - 4y + k = 0

, then the value of

k

are

View answer and explanation

Correct option: A

5, - 35

Option A (

5, - 35

) is correct because these are the values of k for which the line 3x - 4y + k = 0 is tangent to the given circle.

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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