(i) Consider first the case where a^2+b^2\leqslant1, so that p=(a,b)\in S. Then the unique closest point to p is p itself. Now assume that a^2+b^2>1, so that p\notin S. Then the closest point will be the one of the two in the intersection \partial S\cap L of the perimeter \partial S=\{(x,y)\mid x^2+y^2=1\} of the circle and the (unique) line L=\{(x,y)\mid y=bx/a\} which passes through both the origin and p. This point will satisfy x^2+(bx/a)^2=1, i.e. (1+b^2/a^2)x^2=1, i.e. x^2=\frac{1}{1+(b/a)^2}. We choose the square root corresponding to the quadrant of p, i.e. taking it to have the same sign as a. So the unique closest point in S to p is (\pm\sqrt{\frac{1}{1+(b/a)^2}},\frac{b}{a}\frac{1}{1+(b/a)^2}).

(ii) Take S=\varnothing to be the empty set, and any point (a,b).

(iii) Take S=\{(x,y)\mid x^2+y^2=1\} and (a,b)=(0,0).

(iv) The distance is given by \sqrt{|(a-x)^2+(b-mx-c)^2|} = \sqrt{|a^2-2ax+x^2+b^2-2bmx-2bc+2mxc+m^2x^2+c^2|}, which simplifies to \sqrt{|(1+m^2)x^2 + 2(mc-mb-a)x + a^2+b^2+c^2-2bc|}.

The closest point will be that which minimises the distance. We could do some differentiation here to find a minimum point, or we could try something a bit simpler: the closest point will be s\in S such that the line L passing through p and s meets the line S at a right angle. Writing L=\{(x,y)\mid y=m'x+c'\}, this means that we must have m'=-1/m and c'=b-m'a=b+(a/m).

Solving m'x+c'=mx+c then reduces to -(x/m)+b+(a/m)=mx+c, whence (1+m^2)x=mb+a-c, i.e. s=(\frac{mb+a-mc}{1+m^2},m\frac{mb+a-mc}{1+m^2}+c).

N.B. this is really overkill: the question just asks that we prove that a unique point exists, which we could do by just showing that the derivative of the distance function has a unique root corresponding to a minimum. However, this calculation will be useful in the next part of the question.

(v) Note that this point looks exactly like the one we describe at the end of part (iv), with m=2 and c=1, so S is the line given by y=2x+1.

(vi) Assume that there actually are two distinct points of S that are closest to (a,b), and call them p_1=(a_1,b_1) and p_2=(a_2,b_2). Then the line L joining p_1 and p_2 is also contained in S, by hypothesis. But we already know that the closest point on a line to a point is such that the line through it and p meets L at a right angle. This means that the lines L_1 (from p to p_1) and L_2 from (p to p_2) both meet L at a right angle, and so they are parallel. But they must also intersect at p, and this gives a contradiction, since distinct parallel lines in the plane never meet.

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