The equation of the circle passing through the foci of the ellipse x²/16 + y²/9 = 1, and having centre at (0,3) is

The equation of the ellipse is x²/16 + y²/9 = 1.
The foci of the ellipse are at (±ae, 0), where a² = 16, b² = 9, and a² = b² + c².
Thus, c² = a² - b² = 16 - 9 = 7, which means c = √7.
So, the foci are at (±√7, 0).
The circle passes through the foci (±√7, 0) and has a center at (0, 3).
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0.
Since the center is at (0, 3), we have -2g = 0 and -2f = 6, so g = 0 and f = -3.
The equation becomes x² + y² - 6y + c = 0.
Since the circle passes through (±√7, 0), we can substitute these points into the equation:
(√7)² + 0² - 6(0) + c = 0 => 7 + c = 0 => c = -7
(-√7)² + 0² - 6(0) + c = 0 => 7 + c = 0 => c = -7
Therefore, the equation of the circle is x² + y² - 6y - 7 = 0. However, this equation is not among the options. Let's check the options provided.
All given equations of circles have the same center at (0,3). The radius varies. The correct option will be the one that passes through (±√7, 0).