Question regarding optimization using Lagrange multipliers

Question

Let $K=\{(x,y):x^2+y^2=1\}, A=\{(x,y,):2x+3y=10)\}$

Find $x \in K, y\in A$ sucht that $d(x,y)=d(A,K)$.

The lagrangian function is

$$H(x_1,x_2,y_1,y_2,\lambda_1,\lambda_2)=(x_1-y_1)^2+(x_2-y_2)^2+\lambda_1(x_1^2+x_2^2-1)+\lambda_2(2y_1+3y_2-10)$$

The distance of the 2 points would normally be $\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ so obviously we squared it. Why don't we have to do the same for the conditions $x_1^2+x_2^2-1=0$ and $2y_1+3y_2-10=0$?

For example, here, $\ln$ is applied to the condition as well

Answer 1

Lagrangians aren‘t necessarily unique. Optimizing $f(x)$ has the same argmax or argmin as optimizing $\ln(f(x))$ or $f(x)^2$ for $f(x)\in \mathbb{R}_+$. You can also modify constraints. However, notice that satisfying $2y_1 + 3y_2-10=0$ is not the same as satisfying $(2y_1 + 3y_2 - 10)^2=0^2$ for $y\in \mathbb{R}^2$ (but it is the same as satisfying $4y_1 + 6y_2 - 19 = 1$, eg).