Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question 
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$. 
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ that I can maximize?
 A: Although this kind of problem is usually solved using Lagrange multipliers, this one can be solved using methods from precalculus.
We wish to maximize the value of $x^2+y^2$ on the circle
\begin{equation}
(x-2)^2+y^2=1
\end{equation}
Which is defined on the interval $[1,3]$.
Since $x^2+y^2=4x-3$ the maximum and minimum values of $x^2+y^2$ will be the maximum and minimum values of $4x-3$ on the interval $[1,3]$, namely $1$ and $9$.
A: $F(x,y,\lambda)=x^2+y^2+\lambda (x^2-4 x+y^2+3)$ is lagrange multiplier. We need to solve system of equations $$F'_x=0$$ $$F'_y=0,$$ $$x^2-4 x+y^2+3=0$$ or $$2x+2\lambda x-4\lambda=0,$$ $$2y+2\lambda y=0,$$ $$x^2-4 x+y^2+3=0.$$ We have $(x,y,\lambda)\in\{(1,0,1),(3,0,-3)\}$. For $\lambda=1$ is $d^2 F>0$ so $(x,y)=(1,0)$ is point of minimum and $(x,y)=(3,0)$ is point of maximum, since $d^2 F<0$ for $\lambda=-3$. 
A: Hint 
The equation you have is that of a circle,  $(x-2)^2+y^2= 1$, and you want the maximum of $x^2+y^2=4x-3$. Clearly from the latter we need the extreme possible $x$, which from the circle's equation will be when $x\in \{1,3\}$.
A: $x^2 + y^2$ is essentially the square of the distance of a point $(x, y)$ from the origin. Now, the question is asking us to find two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ which lie on the circle and are at a minimum and maximum distance from the origin.
It makes sense that these two points will lie on the line joining the center of the circle and the origin. I won't be proving this here, but it should be pretty obvious upon drawing a rough diagram.
For the given circle
$$
x^2 - 4x + y^2 + 3 = 0
$$
the center is $(2,0)$. Hence, we draw a line passing through this point and the origin (the red line in the below diagram).

The red line is
$$
y = 0
$$
Now we find the point of intersection between this line and the circle.
$$
x^2 - 4x + 3 = 0 \\
(x - 3)(x - 1) = 0 \\
x = 1, 3
$$
Since $y = 0$, the maximum and minimum value of $x^2 + y^2$ is $1$ and $9$.
