JEE MATHEMATICS MCQ: CIRCLES

1 Easy

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Analyzing +4 | -1

P(2,8)

is an interior point of a circle

x^2+y^2-2 x+4 y-\lambda=0

which neither touches nor intersects the axes, then set for

\lambda

View answer and explanation

Correct option: D

\phi

Option D (

\phi

) is correct because there is no value of

\lambda

that satisfies both the condition for P to be an interior point and the condition that the circle neither touches nor intersects the axes.

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{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}},

x

\left( {\alpha ,2{\alpha ^2}} \right)

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{x^2} + {y^2} - 2x + 4y - 4 = 0

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\left( {{x_1},{y_1}} \right)

\quad \alpha, \beta

\gamma

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x^2+y^2=1

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A P, A Q

A R

G P

\cos \alpha / 2, \cos \beta / 2

\cos \gamma / 2

x+y=6

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x+y=5

x-y=1

9 \pi