The largest value of $a$ for which the circle $x^{2} + y^{2} = a^{2}$ falls totally in the interior of the parabola $y^{2} = 4 (x + 4)$ is

4 \sqrt{3}

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4

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4 \frac{\sqrt{6}}{7}

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2 \sqrt{3}

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Solution

The correct option is D $2 \sqrt{3}$
Given equation of parabola is,
$y^{2} = 4 (x + 4)$
Given equation of circle is,
$x^{2} + y^{2} = a^{2}$
$\Rightarrow x^{2} + 4 (x + 4) = a^{2}$
$x^{2} + 4 x + 16 - a^{2} = 0$
For maximum value of $a$ (radius), the circle should touch the parabola.
That means there should be only one solution.
So, $D = 0$
$\Rightarrow 16 - 4 (16 - a^{2}) = 0$
$\Rightarrow a = 2 \sqrt{3}$

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The largest value of a for which the circle x2+y2=a2 falls totally in the interior of the parabola y2=4(x+4) is

The correct option is D 2√3Given equation of parabola is, y2=4(x+4)Given equation of circle is, x2+y2=a2⇒x2+4(x+4)=a2x2+4x+16−a2=0For maximum value of a (radius), the circle should touch the parabola.That means there should be only one solution.So, D=0⇒16−4(16−a2)=0⇒a=2√3

The largest value of $a$ for which the circle $x^{2} + y^{2} = a^{2}$ falls totally in the interior of the parabola $y^{2} = 4 (x + 4)$ is