Cfrgtkky

I'm stuck at this Calculus problem. The question asks to find the closest point on a circle of radius $1$ to the black dot as shown in the figure:

Let $P = (x,y)$ be the point on the unit circle centered at the origin and $Q = (-2,1)$ be the black point you are talking about...then we minimize $PQ^2= (x+2)^2+(y-1)^2= 4x-2y+6$ such that $x^2+y^2=1$. We have $|4x-2y|le sqrt{(4^2+2^2)(x^2+y^2)}=2sqrt{5}implies -2sqrt{5}le 4x-2yle 2sqrt{5}implies PQ_{text{min}}=sqrt{6-2sqrt{5}}=sqrt{5}-1$. The minimum value occurs when $dfrac{4}{x}=dfrac{-2}{y}implies x = -2yimplies P=(x,y) = (-frac{2}{sqrt{5}}, frac{1}{sqrt{5}})$ .

Hint: closest point lies on the line, connecting the center of the circle and this point. Proven by triangle inequality.

Closest point to the circle is the intersection of the line joining the point with the origin and the circle. The equation of line joining the point with the origin $Big($calculated using $y=mx+c$ or $dfrac{y-y_1}{y_2-y_1}=dfrac{x-x_1}{x_2-x_1}$$Big)$ gives you $x=-2y$. So to find the intersection $(-2y)^2+y^2=1implies y=pmdfrac{1}{sqrt{5}}$. So the required point is $left(dfrac{-2}{sqrt{5}},dfrac{1}{sqrt{5}}right)$ and the distance is $sqrt{6-2sqrt{5}}$.