A square is inscribed in the circle x^2+y^2-6x+8y-103=0 with its sides parallel to the coo…

Question

A square is inscribed in the circle $x^2+y^2-6x+8y-103=0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is:

Accepted Answer

Centre $(3,-4)$, $r^2=9+16+103=128\Rightarrow r=8\sqrt2$.

For a square inscribed with sides parallel to the axes, half-side $=\dfrac r{\sqrt2}=8$; vertices at $(3\pm8,-4\pm8)$.

Distances from origin: $\sqrt{137},\sqrt{265},\sqrt{41},13$. The nearest is $\sqrt{41}$ (vertex $(-5,4)$).

Solutions · 1