Question

The equation of the circle passing through the foci of the ellipse `x^2/16 + y^2/9` = 1 and having centre at (0, 3) is

विकल्प

• x² + y² – 6y – 5 = 0

• x² + y² – 6y + 7 = 0

• x² + y² – 6y – 7 = 0

• x² + y² – 6y + 5 = 0

Answer 1

x² + y² – 6y – 7 = 0

Explanation:

Given that equation of ellipse is `x^2/16 + y^2/9` = 1

∴ C² = a² – b² = 16 – 9 = 7 ⇒ C = `+- sqrt(7)` 

∴ Foci of ellipse = `(+-  C, 0) = (+- sqrt(7), 0)`

Since, circle passes through `(+- sqrt(7), 0)` and centre at (0, 3)

∴ `r = sqrt((sqrt(7) - 0)^2 + (0 - 3)^2`

= `sqrt(7 + 9)` = ± 4

Equation of circle is (x – 0)² + (y – 3)² = (± 4)² ⇒ x² + y² – 6y – 7 = 0