JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1

A circle passes through the origin

O

and cuts the axes at

A(a, 0)

and

B(0, b)

. The reflection of origin

O

in the line

A B

is the point

View answer and explanation

Correct option: C

\left(\frac{2 a b}{a^2+b^2}, \frac{2 a b}{a^2+b^2}\right)

The correct answer is B

\left(\frac{{2 a^2 b}}{{{{a^2+b^2}}}}, \frac{{2 a b^2}}{{{{a^2+b^2}}}}\right)

because it represents the reflection of the origin in the line AB, calculated using perpendicularity and equidistance conditions.

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