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Maths-

General

Easy

Question

The radius of the circle passing through the points of intersection of ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction$ = 1 and x²– y² = 0 is -

$\frac{a b}{\sqrt{a^{2} + b^{2}}}$
$\frac{\sqrt{2} a b}{\sqrt{a^{2} + b^{2}}}$
$\frac{a^{2} - b^{2}}{\sqrt{a^{2} + b^{2}}}$
$\frac{a^{2} + b^{2}}{\sqrt{a^{2} - b^{2}}}$

The correct answer is: $fraction numerator square root of 2 a b over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent end root end fraction$

Two curves are symmetrical about both axes and intersect in four points, so, the circle through their points of intersection will have centre at origin.
Solving $x to the power of 2 end exponent minus y to the power of 2 end exponent$ = 0 and $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction$ = 1, we get
$x to the power of 2 end exponent equals y to the power of 2 end exponent$ = $fraction numerator a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction$
Therefore radius of circle
= $square root of fraction numerator 2 a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction end root$ = $fraction numerator square root of 2 a b over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent end root end fraction$

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