Notice that
[tex]x^2+14xy+49y^2=(x+7y)^2[/tex]
so the constraint is a set of two lines,
[tex](x+7y)^2=100\implies\begin{cases}x+7y=10\\x+7y=10\end{cases}[/tex]
and only the first line passes through the first quadrant.
The distance between any point [tex](x,y)[/tex] in the plane is [tex]\sqrt{x^2+y^2}[/tex], but we know that [tex]\sqrt{f(x,y)}[/tex] and [tex]f(x,y)[/tex] share the same critical points, so we need only worry about minimizing [tex]x^2+y^2[/tex]. The Lagrangian for this problem is then
[tex]L(x,y,\lambda)=x^2+y^2+\lambda(x+7y-10)[/tex]
with partial derivatives (set equal to 0)
[tex]L_x=2x+\lambda=0[/tex]
[tex]L_y=2y+7\lambda=0[/tex]
[tex]L_\lambda=x+7y-10=0[/tex]
We have
[tex]L_y-7L_x=2y-14x=0\implies y=7x[/tex]
which tells us that
[tex]x+7y-10=0\iff x+49x=10\implies x=\dfrac15\implies y=\dfrac75[/tex]
so that [tex]\left(\dfrac15,\dfrac75\right)[/tex] is a critical point. The Hessian for the target function [tex]x^2+y^2[/tex] is
[tex]H(x,y)=\begin{bmatrix}2&0\\0&2\end{bmatrix}[/tex]
which is positive definite for all [tex]x,y[/tex], so the critical point is the site of a minimum. The minimum distance itself (which we don't seem to care about for this problem, but we might as well state it) is [tex]\sqrt{\left(\dfrac15\right)^2+\left(\dfrac75\right)^2}=2[/tex].
