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A circle is drawn with the two foci of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
at the end of the diameter. What is the equation of the circle?
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at the end of the diameter. What is the equation of the circle?
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Verified Answer
The correct answer is:
$x^{2}+y^{2}=a^{2}-b^{2}$ (20) 10-I]
Foci of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ are given as $(a e, 0)$ and $(-a, 0)$.
Since, two foci are at the end of the diameter $\therefore$ Equation of circle, is
$(x-a e)(x+a e)+(y-0)(y-0)=0$
$\Rightarrow x^{2}-a^{2} e^{2}+y^{2}=0$
$\Rightarrow x^{2}+y^{2}-a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=0\left(\because e=\sqrt{1-\frac{b^{2}}{a^{2}}}\right)$
$\Rightarrow x^{2}+y^{2}-a^{2}+b^{2}=0$
$\Rightarrow x^{2}+y^{2}=a^{2}-b^{2}$
Since, two foci are at the end of the diameter $\therefore$ Equation of circle, is
$(x-a e)(x+a e)+(y-0)(y-0)=0$
$\Rightarrow x^{2}-a^{2} e^{2}+y^{2}=0$
$\Rightarrow x^{2}+y^{2}-a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=0\left(\because e=\sqrt{1-\frac{b^{2}}{a^{2}}}\right)$
$\Rightarrow x^{2}+y^{2}-a^{2}+b^{2}=0$
$\Rightarrow x^{2}+y^{2}=a^{2}-b^{2}$
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