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

 

 

  • Option 1)

    x^{2}+y^{2}-6y+5=0

  • Option 2)

    x^{2}+y^{2}-6y-7=0

  • Option 3)

    x^{2}+y^{2}-6y+7=0

  • Option 4)

    x^{2}+y^{2}-6y-5=0

 

Answers (2)

As we learnt in

Coordinates of foci -

\pm ae,o

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 

Foci of ellipse are \left ( \pm ae,0 \right )

a^{2}=16;\:b^{2}=9;\:e=\sqrt{1-\frac{b^{2}}{a^{2}}} = \frac{\sqrt{7}}{4}

ae=\sqrt{7}

Circle is 

x^{2}+y^{2}-6y+k=0

It passes through \left ( \sqrt{7},0 \right )

We get, 7+k=0 

\Rightarrow k=-7

x^{2}+y^{2}-6y-7=0


Option 1)

x^{2}+y^{2}-6y+5=0

This option is incorrect.

Option 2)

x^{2}+y^{2}-6y-7=0

This option is correct.

Option 3)

x^{2}+y^{2}-6y+7=0

This option is incorrect.

Option 4)

x^{2}+y^{2}-6y-5=0

This option is incorrect.

N neha

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