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Exercise 5.6 · Q15

Q.The equation of the circle passing through the foci of the ellipse x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}9=1 having centre at (0,3)(0,3) is

(1) x2+y2−6y−7=0x^2+y^2-6y-7=0
(2) x2+y2−6y+7=0x^2+y^2-6y+7=0
(3) x2+y2−6y−5=0x^2+y^2-6y-5=0
(4) x2+y2−6y+5=0x^2+y^2-6y+5=0
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Find the ellipse's foci first (an ordinary c2=a2−b2c^2=a^2-b^2 computation), then this becomes an ordinary centre+point circle problem.

Step 1. Foci of the ellipse. a2=16,b2=9⇒c2=16−9=7⇒c=7a^2=16,b^2=9\Rightarrow c^2=16-9=7\Rightarrow c=\sqrt7. Foci (±7,0)(\pm\sqrt7,0).

Step 2. Radius == distance from centre (0,3)(0,3) to a focus, say (7,0)(\sqrt7,0). …

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