Question

The equation of the circle passing through the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ and having centre at (0, 3) is ________.

• A. $x^{2}+y^{2}-6y-5=0$
• B. $x^{2}+y^{2}-6y+7=0$
• C. $x^{2}+y^{2}-6y-7=0$
• D. $x^{2}+y^{2}-6y+5=0$

Hint:

Standard Circle Equation: $(x-h)^2 + (y-k)^2 = r^2$.

Answer 1

The Correct Option is C

Step 1: Concept
Find foci of the ellipse, then determine the radius of the circle.
Step 2: Analysis of Ellipse
$a^2 = 16$, $b^2 = 9$.
$e = \sqrt{1 - b^2/a^2} = \sqrt{1 - 9/16} = \sqrt{7/16} = \frac{\sqrt{7}}{4}$.
Foci $(\pm ae, 0) = (\pm 4 \cdot \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0)$.
Step 3: Circle Radius
Centre $(h, k) = (0, 3)$.
Radius$^2 = (0 - \sqrt{7})^2 + (3 - 0)^2 = 7 + 9 = 16$.
Step 4: Calculation
Equation: $(x - 0)^2 + (y - 3)^2 = 16$.
$x^2 + y^2 - 6y + 9 = 16 \Rightarrow x^2 + y^2 - 6y - 7 = 0$.
Final Answer:(C)