Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$

Question

Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.

I can't use polar coordinates. I decided to used Lagrange multipliers.

Problem: after using Lagrange multipliers I got this system: $$ \left\{ \begin{array}{c} \frac{x-4}{x}=\frac{y-2}{y} \\ x^2+y^2=a \end{array} \right. $$ And then I add one more equation to the system: $5=(x-4)^2+(y-2)^2$ and $45=(x-4)^2+(y-2)^2$, so I have to find the $a$ that satisfies both system of equations: $$ \left\{ \begin{array}{c} \frac{x-4}{x}=\frac{y-2}{y} \\ x^2+y^2=a \\ 5=(x-4)^2+(y-2)^2 \end{array} \right. $$ and $$ \left\{ \begin{array}{c} \frac{x-4}{x}=\frac{y-2}{y} \\ x^2+y^2=a \\ 45=(x-4)^2+(y-2)^2 \end{array} \right. $$ The thing is that I can't resolve those systems, I already tried a lot of ways but I never get a result. I would really appreciate some help.

Answer 1

Hint

Let $a=b^2$ where $b\ge0$

WLOG any point on $$x^2+y^2=b^2$$ be $P(b\cos t,b\sin t)$

Now $$(b\cos t-4)^2+(b\sin t-2)^2=20+b^2-4b(2\cos t+\sin t)$$

Again $$-\sqrt{2^2+1^2}\le-(2\cos t+\sin t)\le\sqrt{2^2+1^2}$$

Answer 2

The point $(4,2)$ lies on the line $y=\frac{1}{2}x$ which passes through the center of the circle.

Now the minimum and maximum is happened in the two intersection points of the line and the circle.

Answer 3

Perhaps a different approach altogether. If $(4,2)$ is inside the circle of radius $\sqrt{a}$, then the minimum and maximum distances combine into one diameter, hence $$ 2a = d = \sqrt{5} + 3\sqrt{5} = 4\sqrt{5} $$ and the point of interest is $a = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt{5}$ away, which means the point $(4,2)$ cannot be inside the circle.

Then the diameter of the circle is exactly the difference between the distances, i.e. $$ 2a = d = 3\sqrt{5} - \sqrt{5} = 2\sqrt{5}, $$ and so the radius must be $a = \sqrt{5}$.

Answer 4

It can be geometrically seen that once you draw the diameter of the circle that passes through $(4,2)$, contributes the maximum and minimum distances. That diameter must have an equation like $y={x\over2}$ which together with $x^2+y^2=a$, leads to two points $(2\sqrt{a\over 5},\sqrt{a\over 5})$ and $(-2\sqrt{a\over 5},-\sqrt{a\over 5})$. The right choice for $a$ is then straightforward, but the problem may or may not have a solution.