JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Analyzing +4 | -1

Let the centre of a circle

C

(\alpha, \beta)

and its radius

r < 8

. Let

3 x+4 y=24

and

3 x-4 y=32

be two tangents and

4 x+3 y=1

be a normal to

C

. Then

(\alpha-\beta+r)

is equal to

View answer and explanation

Correct option: A

Option A, 7, is correct because solving the equations derived from the tangent and normal conditions gives

\alpha = 1, \beta = -1, r = 5

, and thus

\alpha - \beta + r = 7

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