Jtdylktuy

Find the minimum sum of distances from a point on the circle with equation $x^2+y^2=16$ to the points $A(-2,0)$ and $B(2,0)$.

Does this mean I have to find two distances, one from a point on the circle to point $A$ and one from a point on the circle to point $B$? Or am I supposed to find the distance between all $3$ points? If it's for all $3$, how do I go about doing this? thank you

You are supposed to minimize the sum of $overline{PA}$ and $overline{PB}$ where $P$ is your point on the circle.

You have to set up the system:

Minimize $sqrt{(x+2)^2+y^2}+sqrt{(x-2)^2+y^2}$ under the condition $x^2+y^2=16$

Hint: Use the circle condition to express $x$ by $y$ (or vice versa) in order to eliminate one variable.

Your first sentence is correct. You just need to find two distances. Since you want the minimum, then it will have to be along the line from the point to the center of the circle. The sum of these two distances will then be minimum. Also, these distances will be the distance of the points from the origin minus the radius of the circle.

Consider an ellipse with foci $(-2,0)$, and $(2,0).$

Let the sum of distances of a point $P$ to $(-2,0)$ and $(2,0)$ be $2a.$

Then $b^2= a^2 - 4$, where $b$ is the minor axis, and $a$ the major axis.

The locus of $P$:

$dfrac{x^2}{a^2} +dfrac{y^2}{b^2} =1.$

Minimize $a$ with the restriction:

$P$ is a point on

$x^2+y^2=16.$

$a$ is minimal when the $x=pm$ a, $y=0$. (Why?).

Hence $a^2=16$, $a= 4.$

Recall the sum of distances to to the foci is $2a$ ,

and finally $2a_{min}=8.$