Question

A circle of radius 'r' passes through both foci and exactly four points on the ellipse with equation $x^2+16y^2=16$. The set of all possible values of r is an interval [a,b), then a + b is equal to -

• A. (A) $5\sqrt{2}+4$
• B. (B) $\sqrt{17}+7$
• C. (C) $6\sqrt{2}+3$
• D. (D) $\sqrt{15}+8$

Answer 1

Answer: (D)

$\sqrt{15}+8$

Answer 2

The ellipse equation is $\frac{x^2}{16} + \frac{y^2}{1} = 1$. Here, $a=4$ and $b=1$. The distance from the center to the foci is $c = \sqrt{a^2 - b^2} = \sqrt{16-1} = \sqrt{15}$. The foci are at $(\pm \sqrt{15}, 0)$.

Let the circle have radius $r$ and center $(0, k)$ (since it passes through both foci, its center lies on the y-axis). The distance from $(0, k)$ to $(\sqrt{15}, 0)$ is $r$, so $r^2 = (\sqrt{15}-0)^2 + (0-k)^2 = 15 + k^2$.

The equation of the circle is $x^2 + (y-k)^2 = r^2$. Substituting $x^2 = 16(1-y^2)$ from the ellipse equation: $16(1-y^2) + (y-k)^2 = r^2$ $16 - 16y^2 + y^2 - 2ky + k^2 = 15 + k^2$ $-15y^2 - 2ky + 1 = 0$ $15y^2 + 2ky - 1 = 0$

For exactly four intersection points, the quadratic equation for $y$ must have two distinct real roots, $y_1$ and $y_2$, such that $-1 < y_1 < 1$ and $-1 < y_2 < 1$. This requires:

1. Discriminant $\Delta > 0$: $(2k)^2 - 4(15)(-1) = 4k^2 + 60 > 0$ (always true).
2. $f(1) > 0$ and $f(-1) > 0$, where $f(y) = 15y^2 + 2ky - 1$. $f(1) = 15 + 2k - 1 = 14 + 2k > 0 \implies k > -7$. $f(-1) = 15 - 2k - 1 = 14 - 2k > 0 \implies k < 7$. So, $-7 < k < 7$.

The radius squared is $r^2 = 15 + k^2$. Since $-7 < k < 7$, we have $0 \le k^2 < 49$. Therefore, $15 \le r^2 < 15 + 49$, which means $15 \le r^2 < 64$. Taking the square root (and since $r>0$), we get $\sqrt{15} \le r < 8$. The interval is $[a, b) = [\sqrt{15}, 8)$. So, $a = \sqrt{15}$ and $b = 8$. $a+b = \sqrt{15} + 8$.