Minimize distance between $P$ a point of the circle $x^2+y^2=1$ and $Q=(c,-2c+8)$.

Question

Let $P$ a point of the circle of equation $x^2+y^2=1$ and $Q=(c,-2c+8)$. I have to minimize the distance $$\|P-Q\|=\sqrt{(x-c)^2+(y+2c-8)^2}$$ with the constraint $x^2+y^2-1=0$. I have to use Lagrange Multiplier. I agree that it's equivalent to minimise $(x^2-c)^2+(y+2c-8)^2$ under the constraint $x^2+y^2=1$, but in the solution of my exercise, they say that it's equivalent to minimize $f(x,y)=8-2x-y$ under the constraint $x^2+y^2=0$, but I don't understand why. Any explanation ?

Answer 1

Draw the line $y=-2x+8$.

Let $A(0,8)$, $B(4,0)$, $C(0,0)$ and $CD$ be an altitude of $\Delta ABC$.

Thus, our distance is $$CD-1=\frac{4\cdot8}{\sqrt{4^2+8^2}}-1=\frac{8}{\sqrt5}-1.$$

Let a ray $CD$ intersects our circle in the point $E$.

Let $l$ is a tangent to the circle in $E$.

Since $l\perp CD$, we get that $l$ is parallel to the line $y=-2x+8$

and we need to find the distance between $l$ and the line $y=-2x+8$.

It was explaining of my previous proof.

By the way, a minimum of $f(x,y)=8-2x-y$, where $x^2+y^2=1$

we can find by the following way.

By C-S $$8-2x-y\geq8-\sqrt{(2^2+1^2)(x^2+y^2)}=8-\sqrt5,$$ which is not that we wish.

Another way.

By C-S $$\sqrt{(x-c)^2+(y+2c-8)^2}=\sqrt{x^2-2x+c^2+y^2+2(2c-8)y+(2c-8)^2}=$$ $$=\sqrt{5c^2-32c+65-2(cx+(2c-8)y)}\geq\sqrt{5c^2-32c+65-2\sqrt{(c^2+(2c-8)^2)(x^2+y^2)}}=$$ $$=\sqrt{5c^2-32c+64-2\sqrt{5c^2-32c+64}+1}=$$ $$=\sqrt{5c^2-32c+64}-1=\frac{1}{\sqrt5}\cdot\sqrt{25c^2-160c+320}-1=$$ $$=\frac{1}{\sqrt5}\cdot\sqrt{(5c-16)^2+64}-1\geq\frac{8}{\sqrt5}-1.$$ The equality occurs for $c=3.2$ and $\vec{(c,2c-8)}||\vec{(x,y)},$

which gives the answer: $\frac{8}{\sqrt5}-1$.

Answer 2

The point $Q(c,-2c+8)$ parametrizes the line $L: y=-2x+8$ (i.e. $2x+y-8=0$) .

The distance from the origin $O(0,0)$ to the line $L$ is$$\frac 8{\sqrt{2^2+1^2}}=\frac 8{\sqrt{5}}$$

The circle $C: x^2+y^2=1$ has $O$ as its centre and radius of $1$.

Line $L$ does not intersect circle $C$ as $x^2+y^2=c^2+(-2c+8)^2>1$.

Hence the shortest distance from circle $C$ to line $L$ is $$\color{red}{\frac 8{\sqrt5}-1}$$