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EEJ MAIN Mathematics QUESTION #6042
Question 1

Find the circle passing through the foci of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ and having its centre at $(0,3)$.

  • $x^2+y^2-6y-7=0$✔️
  • $x^2+y^2-6y+7=0$
  • $x^2+y^2-6y-5=0$
  • $x^2+y^2-6y+5=0$
Correct Answer Explanation

For the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$: $a^2=16, b^2=9$, so $c^2=16-9=7$, giving foci at $(\pm\sqrt{7},0)$.

The circle has centre $(0,3)$. Radius $r=$ distance from $(0,3)$ to $(\sqrt{7},0)$:

$r^2=(\sqrt{7}-0)^2+(0-3)^2=7+9=16\Rightarrow r=4$

Equation: $x^2+(y-3)^2=16\Rightarrow x^2+y^2-6y+9=16\Rightarrow x^2+y^2-6y-7=0$