Locus of a midpoint

Question

Let $Γ_1$ be a circle of radius $4$, and let $Γ_2$ be a circle of radius $14$. The distance between the centers of $Γ_1$ and $Γ_2$ is $25$. Let $A$ be a variable point on $Γ_1$, let $B$ be a variable point on $Γ_2$, and let $M$ be the midpoint of $AB$. Let $S$ be the set of all possible locations of $M$. Then find the area of $S$. I am getting $81\pi$. See the figure

With one of my friends, I got this.Let C1 be centered at $(0,0)$ and let C2 be $(25, 0)$. The points that will lie on the boundary of S are 1. Midpoint of (-4, 0) and (11, 0) i.e. (3.5, 0) 2. Midpoint of (4, 0) and (39, 0) i.e. (21.5, 0) 3. Midpoint of the tangents joining C1 and C2

Let the angle that the point of intersection of the tanget to circle C1 and C2 makes with the x axis be $\theta $ then The point of intersection on C1 and C2 are $$(x_1, y_1) = (4\cos{\theta}, 4\sin{\theta})$$ $$(x_2, y_2) = (25+14\cos{\theta}, 14\sin{\theta})$$

The equation of the tangent is $$ y = -\frac{x}{\tan{\theta}} + c$$

Putting the above two points in the line equation and eliminating c gives $$ 10\sin{\theta} = -\frac{25 + 10\cos{\theta}}{\tan{\theta}}$$ $$ 10\sin{\theta}\tan{\theta} = -25 - 10\cos{\theta} \frac{2}{\cos{\theta}} = -5 \cos{\theta} = -\frac{2}{5}$$

This gives $$ \sin{\theta} = \pm \frac{\sqrt{21}}{5}$$

The two pairs of points of intersection in C1 and C2 are $$ (x_1, y_1) = (-\frac{8}{5}, \frac{4\sqrt{21}}{5})$$ $$ (x_2, y_2) = (\frac{97}{5}, \frac{14\sqrt{21}}{5})$$ and $$(x_1, y_1) = (-\frac{8}{5}, -\frac{4\sqrt{21}}{5})$$ $$ (x_2, y_2) = (\frac{97}{5}, -\frac{14\sqrt{21}}{5})$$

This gives the other two mid points as $(\frac{89}{10}, \frac{9\sqrt{21}}{5})$, $(\frac{89}{10}, -\frac{9\sqrt{21}}{5})$

Using all the mid points obtained and putting them into the ellipse equation $$ \frac{(x - x_1)^2}{a^2} + \frac{(y - y_1)^2}{b^2}$$ following are obtained $$ x_1 = \frac{25}{2}$$ and $$ y_1 = 0$$ $ a = 9$ and $b=9$. Where is it wrong.($81\pi$ is wrong!) Please help, thanks.

Answer 1

We can write $$A = A_0 + 2 a (\cos\theta, \sin\theta) \qquad B = B_0 + 2 b (\cos\phi, \sin\phi)$$ where $A_0$ and $B_0$ are the centers of the circles, and $2a$ and $2b$ are the radii. Then $$M = C_0 + (a\cos\theta + b \cos\phi, a\sin\theta+b\sin\phi)$$ where $C_0 = \frac12(A_0+B_0)$. Note that $$\begin{align} |\overline{MC_0}|^2 &= (a\cos\theta+b\cos\phi)^2+(a\sin\theta+b\sin\phi)^2 \\[4pt] &= a^2 + b^2 + 2 a b \cos(\theta-\phi) \end{align}$$ so that $$| a - b | \;\leq\; |\overline{MC_0}| \;\leq\; a+b$$

Therefore, the locus of $M$ is definitely confined to the annulus with outer radius $a+b$ and inner radius $|a-b|$. Once you show that $M$ accounts for all points in that annulus, then the area is $$\pi (a+b)^2 - \pi(a-b)^2 = 4 a b \pi = \pi\cdot(\text{product of radii})$$

Answer 2

it seems an annulus with larger radius 9 and smaller radius 5, and hence area $\pi(9^2-5^2)=56\pi$.

I'll add more details later (I did), but the main idea is, first fix the point $A$ on the left circle and let $B$ vary on the right circle. Then the midpoint $M$ (of $A$ and $B$) itself makes a circle (which I will call a mid-circle) that has radius one half of the radius of the right circle (that is radius $7$).

Say the left circle is centered at the origin and the right circle at $(25,0)$. Fix $A=(4,0)$ and let $B$ vary on the right circle. The corresponding mid-circle intersects the $x$-axis at $(4+39)/2=21.5$, i.e. at $(21.5,0)$, and at $(4+11)/2=7.5$, i.e. at $(7.5,0)$. If we repeat this operation with the point $A=(-4,0)$, then the corresponding mid-circle intersects the $x$-axis at $(3.5,0)$ and at $(17.5,0)$.

Now given any fixed point $A$ on the left circle (and letting $B$ vary on the right circle) the corresponding mid-circle has a "leftmost" point $P$ which is equal to the midpoint of $A$ and $(11,0)$ (the latter being the "leftmost" point of the right circle).

We could think of the locus of these "leftmost" points $P$ as the locus of midpoints of $A$ and $(11,0)$ where $A$ varies on the left circle $\Gamma_1$. This locus is a mid-circle (of the point $(11,0)$ and the left circle $\Gamma_1$), call it $L$, that intersects the $x$-axis at $(-4+11)/2=3.5$, i.e. $(3.5,0)$, and at $(4+11)/2=7.5$, i.e. $(7.5,0)$. Thus $L$ has radius $2$ (one half of the radius of $\Gamma_1$) and center $(5.5,0)$.

Each point of $L$ is the "leftmost" point of a mid-circle determined by a fixed point $A$ on $\Gamma_1$ when we let $B$ vary on $\Gamma_2$. The union of all such mid-circles is the set $S$ of all possible midpoints as defined in the question by the OP. If $P$ is the "leftmost" point of such a mid-circle, then said mid-circle is the translation by $P$ of the circle $J$ of radius $7$ (one half of the radius of the right circle) and center $(7,0)$ (and where the "leftmost" point of $J$ is the origin).

Since $P$ varies on $L$ it follows that the locus $S$ of all midpoints in question as asked by the OP is the same as the Minkowski sum $L+J$. The Minkowski sum of two circles is clearly an annulus (or a disk), and now I need to explain why it is the annulus I think it is ... or to figure which annulus, if my first guess turns out wrong. Working on it.

Clearly this annulus is symmetric about the $x$-axis, and its intersection with the $x$-axis is contained in the line segment with endpoints $(3.5,0)$ and $(21.5,0)$ (and contains these two endpoints), and hence the center of the annulus is $(3.5+21.5)/2=12.5$, that is $(12.5,0)$. So the larger circle of this annulus has radius $21.5-12.5=9$. The annulus also contains the points $(7.5,0)$ and $(17.5,0)$, so the smaller radius is at most $17.5-12.5=5$. This would indeed be exactly the smaller radius provided that there are no points of the annulus that are on the line segment connecting $(7.5,0)$ and $(17.5,0)$ (except for its end-points). The latter looks plausible but needs a proof, let me think.

Well, yes, the curvature of $L$ is bigger that the curvature of $J$ (since the radius of $L$ is $2$ and the radius of $J$ is $7$). It follows that if $P$ is any point on $L$ then the circle $P+J$ (i.e. the translation of $J$ by $P$) intersects the $x$-axis once somewhere to the left of the point $(7.5,0)$, the latter being the "rightmost" point of $L$. (It has a second intersection way-further to the right where I do not care to look, since I find it irrelevant, certainly not between $(7.5,0)$ and $(12.5,0)$.) This indeed implies that the smaller radius of the annulus is $5$, and hence the area of the annulus $S$ is $\pi(9^2-5^2)=\pi(81-25)=56\pi$.

Answer 3

This is most likely analytic geometry, so I'd say we can put:

$$\Gamma_1\;:\;\; x^2+y^2=16\;\;;\;\;\;\Gamma_2\;:\;\; (x-25)^2+y^2=196$$

Take now for example

$$\begin{cases}A=\left(x\,,\,\sqrt{16-x^2}\right)\in\Gamma_1&,\;-4\le x\le4\\{}\\B'=\left(11\,,\,0\right)\in\Gamma_2\end{cases}$$

The midpoint of $\;AB'\;$ is given by

$$\left(\,\frac{x+11}2\;,\;\frac{\sqrt{16-x^2}}2\,\right)$$

We now have

$$\left(\frac{x+11}2-\frac{11}2\right)^2+\frac{16-x^2}4=\frac{x^2}{4}+4-\frac{x^2}4=4$$

and we thus get the circle $\;\left(x-\frac{11}2\right)^2+y^2=4\;$ upon taking both signs in the $\;y\;$-coordinate of $\;A\;$

Try to generalize the above to any point $\;B\in\Gamma_2\;$ .

Answer 4

We need only to consider midpoint of rays through origin for direct (not inverse) tangent to trace required locus, here shown as outer perimeter of yellow circle.The radius of loci of midpoints appears to be average of tangent circles $ (4+14)/2 =9 $ with area $ \pi\cdot 9^2 = 81 \pi,$ for me also.

The above geometrical arrangement has a bit of intuition in-built, so it should be removed to establish result by direct formulation as follows:

Since points on given circles are arbitrary,I assigned parameters $ u,v $ for rotation about their centers. You want to see excursion of midpoint in the whole field of variation of $ u,v.$ which in fact is a surface we are seeking which must be bound by this parametrization.

It is convenient to choose meeting point of direct tangents as origin/pole as we want midpoints of extremal positions on circle.

To find position of center of smaller circle with respect to origin/pole we apply triangle similarity relation:

$ \dfrac{h_1}{4}=\dfrac{h_1+ 25}{14}, \rightarrow ( h_1= 10,h_2= 35) $

Given circles all general points are:

$$ x_1 = a \cos u + h_1, y_1 = a \sin u ; x_2 = b \cos v + h_1+h_2, y_2 = b \sin v ; $$

Midpoint as function of parameters $u,v$

$$ (x,y) = [(x_1+ x_2)/2 , (y_1+ y_2)/2 ] = ( a \cos u + b \cos v + h1 + h2)/2, (a \sin u + b \sin v)/2; $$

When $ v = 0, u =0,$ we get respectively two curves which are readily recognized as circles:

$$ ( x - \frac{b+ h_1+h_2}{2})^2 + y^2 = (a/2)^2, $$

$$( x - \frac{a +h_1+h_2}{2})^2 + y^2 = (b/2)^2 $$

with radii $ (a/2,b/2) $ and $h $ displacements on x-axis shown in thick black color as:

$$ ( \frac{b+ h_1+h_2}{2},\frac{a+ h_1+h_2}{2} ) $$ respectively touching at their starting point parameters $ u = v = 0 ,$ which gives radius of circle S required to be $(a + b)/2 = (14+4)/2 = 9. $ Its area $\pi \, 9^2 = 81 \pi $.

( The inner circle has radius $(a - b)/2 $ for another locus but not relevant here.The included tangent contact circle of annulus has a radius: $ (a-b)/4 $).

Had it not been a circle,envelope may have to be found as a singular solution.

The sketch incorporates all given constants.

EDIT1:

Only as a hindsight I could recognize the radius of contacting circle S at mid-point ought to be directly arrived at $ (a+b)/2$, after fixing location of pole !