The tangent to the circle $x^2+y^2=5$ at $(1,-2)$ touches the circle $x^2+y^2-8 x+6 y+20=0$ at the point

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

The tangent to the circle

x^2+y^2=5

(1,-2)

touches the circle

x^2+y^2-8 x+6 y+20=0

at the point

View answer and explanation

Correct option: B

(3,-1)

Option B ((3,-1)) is correct because the tangent to the first circle at (1,-2) is also tangent to the second circle at (3,-1).

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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