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

Options

A. `x^(2)+Y^(2)-6y =7=0`
B. `x^(2)+Y^(2)-6y +7=0`
C. `x^(2)+Y^(2)-6y -5=0`
D. `x^(2)+Y^(2)-6y -5=0`

Answer

A

Explanation

for the ellipse `(x^(2))/(16)+(y^(2))/(9)=1` , we have a=4 and b=3
`therefore e^(2)=1-(b^(2))/(a^(2))=1-(9)/(16)=(7)/(16)`
`implies e=(sqrt(7))/(4)`
the coordinates of the foci are `S(sqrt(7(,0) and S' (-sqrt(7),0)`
the coordinates of the centre X are (0,3)
Let r be radius of the circle . then ,
`r=CS =sqrt(((0-sqrt(7))^(2))+(3-0)^(2))=4`
hence , the equation of the circle is `(x-0)^(2)+(Y-3)^(2)=4^(2)`
`or , x^(2)+y^(2)-6y -7=0`