JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

Let the circle

C_1: x^2+y^2-2(x+y)+1=0

and

C_2

be a circle having centre at

(-1,0)

and radius 2. If the line of the common chord of

{C}_1

and

{C}_2

intersects the

\mathrm{y}

-axis at the point

{P},

then the square of the distance of

{P}

from the centre of

{C}_1

is:

View answer and explanation

Correct option: A

Option A (2) is correct because the square of the distance is calculated using the coordinates of point P and the center of circle C1.

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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