jmc

We can assume that the ellipse is tangent to the circle $(x-1{)}^2+y^2=1.$ From this equation, $y^2=1-(x-1{)}^2.$ Substituting into the equation of the ellipse, we get $$\frac{x^2}{a^2}+\frac{1-(x-1{)}^2}{b^2}=1.$$This simplifies to $$(a^2-b^2)x^2-2a^{2}x+a^2{b}^2=0.$$By symmetry, the $x$-coordinates of both tangent points will be equal, so the discriminant of this quadratic will be 0: $$(2a^2{)}^2-4(a^2-b^2)(a^2{b}^2)=0.$$This simplifies to $a^4{b}^2=a^4+a^2{b}^4.$ We can divide both sides by $a^2$ to get $$a^2{b}^2=a^2+b^4.$$Then $$a^2=\frac{b^4}{b^2-1}.$$The area of the ellipse is $\pi ab.$ Minimizing this is equivalent to minimizing $ab,$ which in turn is equivalent to minimizing $$a^2{b}^2=\frac{b^6}{b^2-1}.$$Let $t=b^2,$ so $$\frac{b^6}{b^2-1}=\frac{t^3}{t-1}.$$Then let $u=t-1.$ Then $t=u+1,$ so $$\frac{t^3}{t-1}=\frac{(u+1{)}^3}{u}=u^2+3u+3+\frac{1}{u}.$$By AM-GM, $$\begin{array}{rl}{u}^2+3u+\frac{1}{u}& =u^2+\frac{u}{2}+\frac{u}{2}+\frac{u}{2}+\frac{u}{2}+\frac{u}{2}+\frac{u}{2}+\frac{1}{8u}+\frac{1}{8u}+\frac{1}{8u}+\frac{1}{8u}+\frac{1}{8u}+\frac{1}{8u}+\frac{1}{8u}+\frac{1}{8u}\\ & \ge 15\sqrt{u^2\cdot \frac{u^6}{2^6}\cdot \frac{1}{8^8{u}^8}}=\frac{15}{4}.\end{array}$$Equality occurs when $u=\frac{1}{2}.$ For this value of $u,$ $t=\frac{3}{2},$ $b=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2},$ and $a=\frac{3\sqrt{2}}{2}.$ Hence, $$k=ab=\boxed{\frac{3\sqrt{3}}{2}}.$$