Ellipse

\hspace{-16}$Given $\bf{x^2+4y^2=16}$ and form a circle whose center is at $\bf{(1,0)}$\\\\\\ Let we assume that radius of circle is $\bf{=a}$, Then equation is $\bf{(x-1)^2+y^2=a^2}$\\\\\\ So if the circle has largest radius $\bf{a}$. Then circle and ellipse touch each\\\\\\ Other. So condition of tangency is achieved.\\\\\\ Now we solve $\bf{\begin{vmatrix} \bf{x^2+4y^2=16} & \\\\ \bf{(x-1)^2+y^2=a^2}& \end{vmatrix}}$\\\\\\ We get $\bf{3x^2-8x-4(a^2-5) =0 }$\\\\\\ Now for Condition of Tangency,Two curves touch.\\\\\\ So Given equation has equal root.\\\\\\ So $\bf{D=0\Rightarrow 64+4\cdot 3.\cdot4(a^2-5) = 0}$\\\\\\ So $\bf{16\cdot \left(4+3a^2-15\right)=0\Rightarrow a = \pm \sqrt{\frac{11}{3}}}$\\\\\\ So radius $\bf{\left|a\right| = \sqrt{\frac{11}{3}}}.$\\\\\\So equation of Largest circle is $\bf{(x-1)^2+y^2=a^2=\frac{11}{3}}$