JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1

The point on the circle

x^2+y^2=a^2

in first quadrant, so that tangent drawn at this point make a triangle of area

a^2

with the coordinate axes, is

View answer and explanation

Correct option: B

\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)

Option B (

\left(\frac{{a}}{{\sqrt{{2}}}}, \frac{{a}}{{\sqrt{{2}}}}\right)

) is correct because the area of the triangle formed by the tangent and the axes is

a^2

, which leads to

\theta = \frac{{\pi}}{{4}}

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