Explanation:
B We have, Equation of circle -
\((x-1)^2+y^2=r^2\)
\(\{\) Centre \((1,0) \&\) radius \(=r\}\)
From equation (i) with ellipse equation, we get-
\(x^2+4 y^2=16\)
\(x^2+4\left[r^2-(x-1)^2\right]=16\)
\(x^2+4\left(r^2-x^2-1+2 x\right)=16\)
\(x^2+4 r^2-4 x^2-4+8 x=16\)
\(-3 x^2+8 x+4 r^2-20=0\)
\(3 x^2-8 x+20-4 r^2=0\)
Therefore, circle is inscribed in the ellipse So, circle touches the ellipse Hence, equation (ii), we get-
\((-8)^2-4 \times 3 \times\left(20-4 r^2\right)=0\)
\(4-3(5-r)=0\)
\(4-15+3 r^2=0\)
\(3 r^2=11\)
\(r^2=\frac{11}{3}, \quad r=\sqrt{\frac{11}{3}}\)
\(\mathrm{r}^2=\frac{11}{3}, \mathrm{r}=\sqrt{\frac{11}{3}}\)