Consider a family of nested ellipses with foci at points A and B. Each such ellipse represents a set of locations on the plane such that the sum of distances to the foci is constant. As we increase the size of the ellipse it will eventually touch the circle; this is the solution we are looking for. It makes sense to choose the coordinates such that the two foci are at points (a,0) and (-a,0) where $a=|AB|/2$. Let the family of nested ellipses be represented by parameter h which is the length of its shorter semi-axis. Then the equation of such an ellipse is
$ \frac{x^2}{l^2} + \frac{y^2}{h^2} = 1$
Here $l=l(h)$ is the longer semi-axis, which is found from the condition $2 \sqrt{a^2+h^2} = (l-a)+(l+a)$, so $l=\sqrt{h^2+a^2}$ and the ellipse equation is
$ \frac{x^2}{h^2+a^2} + \frac{y^2}{h^2} = 1$
The circle is of course represented by
$(x-x_o)^2 + (y-y_o)^2 = c^2$
Now we have a system of two algebraic equations and the problem boils down to determining when they have a single solution which is the touch point. Note that there are two cases of touching - touching on the near side of the circle, and touching on the far side when the circle is inside of the ellipse. One can eliminate one of the variables from the equations which will convert the problem to a quartic equation in one variable; so a closed form solution can be found by a known formula. A certain value of parameter h will coalesce the two real roots into one, that's the condition of touching.
From the above considerations one can re-formulate the problem in purely geometrical terms: The point on the circle X minimizing the sum of distances |AX|+|BX| satisfies the condition that the line (OX) is the bisector of angle AXB. This formulation leads to an alternative approach to solving the problem which can be pursued algebraically, using trigonometric relations; but perhaps there is some clever geometric shortcut. An approximate solution can be found using geometric optics: Assume the circle is a convex mirror and find an image A' of point A. The image will be virtual (inside the circle), and the line A'B will intersect the circle at point X satisfying the reflection conditions and minimizing the sum of distances |AX|+|BX|. However this geometric optic solution will work only as long as the circle can be approximated by a parabola (paraxial approximation) which is satisfied in the limit of $\angle$AOB $\ll$ 1.
P.S. It looks like a general geometric optics solution (without approximations) is worked out in www.geometrictools.com/Documentation/SphereReflections.pdf and it leads to a quartic equation.
Let consider for $t\in[0,1]$, then
- $A+t(B-A)$ is the line from $A$ to $B$
the mid point $M$ between $A$ and $B$ is
finally we need to add a term to reach $C$, that is for example
$$\operatorname{curve}(t)=A+t(B-A)+4t(t-1)(C-M)$$
with
- $\operatorname{curve}(0)=A$
- $\operatorname{curve}(1/2)=C$
- $\operatorname{curve}(1)=B$
Best Answer
The equation of a line is $y = mx + b$. Where $m$ is the slope.
In this line $m = \frac{200 - 100}{200-100} = \frac{100}{100} = 1$
So $y = x + b$. We can solve for $b$ by plugging in either $(100,100)$ or $(200,200)$ into the equation to get $200 = 200 + b$ or $100 = 100 + b$ so $b = 0$.
So the equation is $y = x$.
Pick a point $(x',y'); y' = x'$ on the line. If we extend the line so that the $x$ value increases by $h$ then what does the $y$ value increase by? $y' = x'$ so $y_{new} = x' + h = y' + h$. So the $y$ value also increases by $h$.
So what does the total distance increase by? $D((x',y'),(x'+h,y'+h)) = \sqrt{(x'+h - x')^2 + (y'+h - y')^2} = \sqrt{h^2 + h^2} = \sqrt{2h^2} = h\sqrt{2}$.
So if you want to increase the line by $D$ you must extend $x$ (and $y$) by... $D = h\sqrt2 \implies h = D/\sqrt2 = D*\frac{\sqrt 2}{2}$.
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So for instance, if we want to extend the line $75$ units, we must increase $x$ by $75*\frac{\sqrt 2}{2}$ units to the point $(x = 100 +75*\frac{\sqrt 2}{2}, y = 100 +75*\frac{\sqrt 2}{2})$.