Let the equation of an ellipse be `(x^2)/(144)+(y^2)/(25)=1`,Then , the radius of the circle with centre `(0,sqrt(2))` and passing through the foci of the ellipse is

Options

A. 9
B. 7
C. 11
D. 5

Answer

C

Explanation

Given equation of ellipse is `(x^2)/(144)+(y^2)/(25)=1`
Here ,`a^2=144 and b^2=25`
Now , `e=sqrt(1-(b^2)/(a^2))=sqrt(1-(25)/(144))`
`=sqrt((119)/(144))=(sqrt(119))/(12)`
`therefore` Foci of an ellipse `=(+-ae,0)`
`=(+- 12 xx(sqrt(119))/(12),0)=(+-sqrt(119),0)`
Since , the circle with centre `(0sqrt(2))` and passing through foci `(+-sqrt(119),0)` of the ellipse.
`therefore` Radius of circle `=sqrt((sqrt(119)-0)^2+(0-sqrt(2))^2)`
`=sqrt(119+2)=sqrt(121)=11`