Q120 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
The radius of the circle passes through the foci of a conic $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ and has its centre at $(0,3)$, then the diameter of the circle is ---
- Correct answerA8 units
- B4 units
- C7 units
- D$2 \sqrt{12}$ units
Solution
The given equation of the ellipse is $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.
Here $a^2 = 16$ and $b^2 = 9$. The eccentricity $e$ is given by $e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{9}{16}} = \sqrt{\dfrac{7}{16}} = \dfrac{\sqrt{7}}{4}$.
The foci of the ellipse are $(\pm ae, 0)$. Substituting the values, $ae = 4 \times \dfrac{\sqrt{7}}{4} = \sqrt{7}$. Thus, the foci are $F_1 = (\sqrt{7}, 0)$ and $F_2 = (-\sqrt{7}, 0)$.
The circle passes through these foci and has its center at $C = (0, 3)$. The radius $r$ of the circle is the distance between the center $(0, 3)$ and either focus $(\sqrt{7}, 0)$.
$r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4$.
The diameter of the circle is $d = 2r = 2 \times 4 = 8$ units.
Answer: 8 units
Here $a^2 = 16$ and $b^2 = 9$. The eccentricity $e$ is given by $e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{9}{16}} = \sqrt{\dfrac{7}{16}} = \dfrac{\sqrt{7}}{4}$.
The foci of the ellipse are $(\pm ae, 0)$. Substituting the values, $ae = 4 \times \dfrac{\sqrt{7}}{4} = \sqrt{7}$. Thus, the foci are $F_1 = (\sqrt{7}, 0)$ and $F_2 = (-\sqrt{7}, 0)$.
The circle passes through these foci and has its center at $C = (0, 3)$. The radius $r$ of the circle is the distance between the center $(0, 3)$ and either focus $(\sqrt{7}, 0)$.
$r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4$.
The diameter of the circle is $d = 2r = 2 \times 4 = 8$ units.
Answer: 8 units
Hence, the correct answer is option: 'A'.