Question

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

• A. $4\sqrt{3}$
• B. $4$
• C. $4\sqrt{\frac{6}{7}}$
• D. $2\sqrt{3}$

Answer 1

The correct option is D $2\sqrt{3}$

Given that the circle $x^2+y^2=a^2$ falls entirely inside the parabola $y^2=4(x+4)$
$\Rightarrow$ Parabola touches the circle externally.
substituting $y^2=4(x+4)$ in $x^2+y^2=a^2$ gives $x^2+4x+16-a^2=0$
Roots of the above quadratic should be real and equal.
$\Rightarrow D=0$
$\Rightarrow 16=4(16-a^2)$
$\Rightarrow a^2=12$
$\therefore$ Largest possible values of $a$ for which circle falls inside parabola is $a=2\sqrt{3}$
Hence, option D.