JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Analyzing +4 | -1

Let

A(1,\,4)

and

B\left( {1,\, - 5} \right)

be two points. Let

P

be a point on the circle

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

, such that

{\left( {PA} \right)^2} + {\left( {PB} \right)^2}

have maximum value, then the points

P

A

and

B

lie on

View answer and explanation

Correct option: A

a straight line

Option A (a straight line) is correct because the points P, A, and B are collinear when (PA)² + (PB)² is maximized, resulting in a straight line.

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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

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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

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x+y=5

x-y=1

9 \pi