Question

Find a point on the curve $x^2 + y^2 = 6$ whose distance from the line $x + y = 7$, is as small as possible.

Answer 1

The point is $(\sqrt{3}, \sqrt{3})$.

Answer 2

The distance from a point $(x, y)$ on the circle $x^2+y^2=6$ to the line $x+y-7=0$ is $D = \frac{|x+y-7|}{\sqrt{2}}$. To minimize $D$, we must minimize $|x+y-7|$. Since the maximum value of $x+y$ on the circle is $2\sqrt{3} < 7$, the expression $x+y-7$ is always negative. Thus, $|x+y-7| = -(x+y-7) = 7-(x+y)$. Minimizing $D$ is equivalent to minimizing $7-(x+y)$, which means maximizing $x+y$. The point on the circle closest to the line occurs where the tangent is parallel to the line $x+y=7$. The slope of the line is $-1$. The slope of the tangent to the circle $x^2+y^2=6$ is $\frac{dy}{dx} = -\frac{x}{y}$. Setting $-\frac{x}{y} = -1$ yields $x=y$. Substituting $x=y$ into $x^2+y^2=6$ gives $2x^2=6$, so $x^2=3$, which means $x = \pm\sqrt{3}$. The possible points are $(\sqrt{3}, \sqrt{3})$ and $(-\sqrt{3}, -\sqrt{3})$. To maximize $x+y$, we choose the point $(\sqrt{3}, \sqrt{3})$, where $x+y = 2\sqrt{3}$.