Question

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

Correct Answer - A
given equation of ellipse of ellipse is `(x^(2))/(16)+(y^(2))/(9)=1`

Here` a=4,b=3, e=sqrt(1-(9)/(16))implies (sqrt(7))/(4)`
` therefore `Foci `=(+_4xx(sqrt(7))/(4),0)=(+_sqrt(7),0)`
Radius of the circle is , `r= sqrt((ae)^(2)+b^(2))`
`=sqrt(7+9)=sqrt(16)=4`
NOw , equation of circle is
`(x-o)^(2)+(y-3)^(2)=16`
`therefore x^(2)+y^(2)-6y-7=0`