The length of the tangent drawn from the point $(4, - 1)$ to the circle $2 x^{2} + 2 y^{2} = 1$ is :

\sqrt{\frac{17}{2}}

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\sqrt{33}

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\sqrt{\frac{33}{2}}

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\sqrt{2}

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Solution

The correct option is D $\sqrt{\frac{33}{2}}$
REF.Image
center = (0,0)
and
radius = $\frac{1}{\sqrt{2}}$
AT = length of tangent
$O T = \sqrt{(4 - 0)^{2} + (- 1 - 0)^{2}}$
$= \sqrt{17}$
$∴ A T = \sqrt{(O T)^{2} - (O A)^{2}}$
$= \sqrt{17 - \frac{1}{2}} = \sqrt{\frac{33}{2}}$
Answer.

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The length of the tangent drawn from the point (4,−1) to the circle 2x2+2y2=1 is :

The correct option is D √332REF.Imagecenter = (0,0)andradius = 1√2AT = length of tangentOT=√(4−0)2+(−1−0)2=√17∴AT=√(OT)2−(OA)2=√17−12=√332Answer.

The length of the tangent drawn from the point $(4, - 1)$ to the circle $2 x^{2} + 2 y^{2} = 1$ is :

The correct option is D $\sqrt{\frac{33}{2}}$
REF.Image
center = (0,0)
and
radius = $\frac{1}{\sqrt{2}}$
AT = length of tangent
$O T = \sqrt{(4 - 0)^{2} + (- 1 - 0)^{2}}$
$= \sqrt{17}$
$∴ A T = \sqrt{(O T)^{2} - (O A)^{2}}$
$= \sqrt{17 - \frac{1}{2}} = \sqrt{\frac{33}{2}}$
Answer.