JEE Main · 2012 · 29 Apr 2012 · Offline · Ellipse

The circle $(x-1{)}^2+y^2=1$ has radius $1$, so its diameter $2$ serves as the semi-minor axis, $b=2$. The circle $x^2+(y-2{)}^2=4$ has radius $2$, so its diameter $4$ serves as the semi-major axis, $a=4$. Centred at the origin with axes along the coordinate axes, the ellipse is $\frac{x^2}{16}+\frac{y^2}{4}=1$, which on clearing denominators becomes $x^2+4y^2=16$.