Given the center and radius of a circle, and one axis of an ellipse, find the value of the other axis of the ellipse that lies tangent to the circle.

Question

Find b.

Given the center and radius of a circle, and one axis of an ellipse, find the value of the other axis of the ellipse that lies tangent to the circle.

$$(x - h)^2 + (y - k)^2 = r^2$$

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Also, I feel compelled to add this edit due to the magnificent tedium and confusion I have experienced in attempting to decipher some of the answers:

I have already solved this problem! I simply mapped the line of all $y$-values for $b$ of the $x$-value of the center of the circle on the offset curve of the ellipse (offset by the radius, of course), then I drew the curve of those values with $b$ on the $x$-axis, and $y$ on the $y$-axis, only to discern the $x$-value of the point of crossing of the horizon extending from the circle's center.

And so, I am not here to have the years eroded away by riddles, nor am I seeking the most lengthy and complicated answer; quite the contrary! I only hope to find the most concise and simple way to solve this particular problem.

There seem to be at least two fairly simple methods:

1. Solving the system of equations

2. The dual-curve crossing

Please dismiss the given numerical values! I only provided them because I assumed that it would make the problem easier to understand. I was wrong. So, please only use the letter variables, as that turns out to be much less confusing.

I have updated the graphic to include the derivative(s) $\frac{dy}{dx}$ and their perpendiclar(s) $m$. When I first posted this problem, I thought someone devised a perhaps even simpler solution involving only algebra using the slopes. Do re-post!

Answer 1

We find the coordinates of point of tangency. At this point the derivatives of both functions must be equal:

$(x-h)^2+(y-k)^2=r^2\rightarrow y'=\frac{-x+h}{\sqrt{r^2-(x-h)^2}}$

$\frac {x^2}{a^2}+\frac{y^2}{b^2}=1 \rightarrow y'=\frac{-bx}{a\sqrt{a^2-x^2}}$

This point must also satisfy both equations,

$\begin{cases}\frac{-bx}{a\sqrt{a^2-x^2}}=\frac{-x+h}{\sqrt{r^2-(x-h)^2}}\\y=\frac ba\sqrt{a^2-x^2}=k+\sqrt{r^2-(x-h)^2}\end{cases}$

that is we have following system to solve(for given values):

$\begin{cases}\frac{-bx}{2\sqrt{4-x^2}}=\frac{-x+1}{\sqrt{1-(x-1)^2}}\\y=\frac b2\sqrt{4-x^2}=5+\sqrt{1-(x-1)^2}\end{cases}$

We may write:

$$\frac{\sqrt{2x-x^2}}{1-x}+5=\frac{2\sqrt{4-x^2}}{-bx}+5$$

$$\frac{\sqrt{2x-x^2}+5-5x}{1-x}=\frac{2\sqrt{4-x^2}}{-bx}+5$$

$\Rightarrow$ $$\frac{y-5x}{1-x}+\frac 1{xy}-5=0$$

$$10x^2y+xy^2-5xy+1=0$$

and gives this system to solve:

$$\begin{cases}10x^2y+xy^2-5xy+1=0\\y=5+\sqrt{2x-x^2}\end{cases}$$

which gives $x\approx \frac 12$, $y\approx 4$ and finally gives $b=2\sqrt5\approx 4.47$

Answer 2

Given $x^2/a^2+y^2/b^2=1,$ find the values of $b>0$ that gives an ellipse tangent to $(x-h)^2+(y-k)^2=r^2.$

R=QQ[x,y,a,b,h,k,r,MonomialOrder=>Eliminate 2]
I=ideal((h*x+k*y+1)^2-r^2*(x^2+y^2),-a^2*x^2-b^2*y^2+1,determinant matrix{{diff(x,(h*x+k*y+1)^2-r^2*(x^2+y^2)),diff(x,-a^2*x^2-b^2*y^2+1)},{diff(y,(h*x+k*y+1)^2-r^2*(x^2+y^2)),diff(y,-a^2*x^2-b^2*y^2+1)}})

$$(a^4r^2-2a^2h^2r^2+h^4r^2-2a^2r^4-2h^2r^4+r^6)b^8+(-2a^6r^2+6a^4h^2r^2-6a^2h^4r^2+2h^6r^2-4a^4k^2r^2+6a^2h^2k^2r^2-2h^4k^2r^2+2a^4r^4+4a^2h^2r^4-6h^4r^4+6a^2k^2r^4+4h^2k^2r^4+2a^2r^6+6h^2r^6-2k^2r^6-2r^8)b^6+(a^8r^2-4a^6h^2r^2+6a^4h^4r^2-4a^2h^6r^2+h^8r^2+6a^6k^2r^2-10a^4h^2k^2r^2+2a^2h^4k^2r^2+2h^6k^2r^2+6a^4k^4r^2-6a^2h^2k^4r^2+h^4k^4r^2+2a^6r^4-8a^4h^2r^4+10a^2h^4r^4-4h^6r^4-8a^4k^2r^4-6a^2h^2k^2r^4-6h^4k^2r^4-6a^2k^4r^4-2h^2k^4r^4-6a^4r^6-8a^2h^2r^6+6h^4r^6+4a^2k^2r^6+6h^2k^2r^6+k^4r^6+2a^2r^8-4h^2r^8-2k^2r^8+r^{10})b^4+(-2a^8k^2r^2+6a^6h^2k^2r^2-6a^4h^4k^2r^2+2a^2h^6k^2r^2-6a^6k^4r^2+2a^4h^2k^4r^2+4a^2h^4k^4r^2-4a^4k^6r^2+2a^2h^2k^6r^2-2a^8r^4+6a^6h^2r^4-6a^4h^4r^4+2a^2h^6r^4+4a^6k^2r^4-6a^4h^2k^2r^4+2a^2h^4k^2r^4+10a^4k^4r^4+2a^2h^2k^4r^4+2a^2k^6r^4+2a^6r^6+4a^4h^2r^6-6a^2h^4r^6-8a^4k^2r^6-10a^2h^2k^2r^6-6a^2k^4r^6+2a^4r^8+6a^2h^2r^8+6a^2k^2r^8-2a^2r^{10})b^2+a^8k^4r^2-2a^6h^2k^4r^2+a^4h^4k^4r^2+2a^6k^6r^2+2a^4h^2k^6r^2+a^4k^8r^2-2a^8k^2r^4+4a^6h^2k^2r^4-2a^4h^4k^2r^4-6a^6k^4r^4-6a^4h^2k^4r^4-4a^4k^6r^4+a^8r^6-2a^6h^2r^6+a^4h^4r^6+6a^6k^2r^6+6a^4h^2k^2r^6+6a^4k^4r^6-2a^6r^8-2a^4h^2r^8-4a^4k^2r^8+a^4r^{10}$$

Then solve this quartic in $b^2.$

Older answers

Given $x^2/2^2+y^2/b^2=1,$ find the values of $b>0$ that gives an ellipse tangent to $(x-1)^2+(y-5)^2=1.$

Since $x=2$ is an outer common tangent for all $b$ there's a unique solution.

We can find the tact invariant as in 3264 conics in a second in M2 with $c=\frac1{b}:$

R=QQ[x,y,c,MonomialOrder=>Eliminate 2]
I=ideal(x^2/2^2+c^2*y^2-1,(x-1)^2+(y-5)^2-1,determinant matrix{{diff(x,x^2/2^2+c^2*y^2-1),diff(x,(x-1)^2+(y-5)^2-1)},{diff(y,x^2/2^2+c^2*y^2-1),diff(y,(x-1)^2+(y-5)^2-1)}})
gens gb I -- matrix {{19008*c^8-2187*c^6+81*c^4-c^2, 65*y-7489152*c^6+760302*c^4-18138*c^2-325, 13*x-1406592*c^6+98478*c^4-816*c^2-26}}

$19008c^8-2187c^6+81c^4-c^2=0.$

$c \approx 0.2342238546756076$

$b \approx 1/0.2342238546756076\approx 4.269419958889189$

Note $c=0, b=\infty$ i.e. the line pair $x^2=2^2$ comes out as a solution, but is not an ellipse.

This method also solves the more general

Given $x^2/2^2+y^2/b^2=1,$ find the values of $b>0$ that gives an ellipse tangent to $(x-h)^2+(y-k)^2=r^2:$

$(16h^4k^4+32h^2k^6+16k^8-32h^4k^2r^2-96h^2k^4r^2-64k^6r^2+16h^4r^4+96h^2k^2r^4+96k^4r^4-32h^2r^6-64k^2r^6+16r^8-128h^2k^4+128k^6+256h^2k^2r^2-384k^4r^2-128h^2r^4+384k^2r^4-128r^6+256k^4-512k^2r^2+256r^4)c^8+(8h^6k^2+16h^4k^4+8h^2k^6+8h^6r^2+8h^4k^2r^2+8h^2k^4r^2+8k^6r^2-24h^4r^4-40h^2k^2r^4-24k^4r^4+24h^2r^6+24k^2r^6-8r^8-96h^4k^2+32h^2k^4-64k^6-96h^4r^2-96h^2k^2r^2+160k^4r^2+64h^2r^4-128k^2r^4+32r^6+384h^2k^2-384k^4+384h^2r^2+256k^2r^2+128r^4-512k^2-512r^2)c^6+(h^8+2h^6k^2+h^4k^4-4h^6r^2-6h^4k^2r^2-2h^2k^4r^2+6h^4r^4+6h^2k^2r^4+k^4r^4-4h^2r^6-2k^2r^6+r^8-16h^6+8h^4k^2-24h^2k^4+40h^4r^2-24h^2k^2r^2-24k^4r^2-32h^2r^4+16k^2r^4+8r^6+96h^4-160h^2k^2+96k^4-128h^2r^2-128k^2r^2-96r^4-256h^2+384k^2+128r^2+256)c^4+(2h^6-2h^4k^2-6h^4r^2+4h^2k^2r^2+6h^2r^4-2k^2r^4-2r^6-24h^4+24h^2k^2+16h^2r^2+24k^2r^2+8r^4+96h^2-64k^2+32r^2-128)c^2+h^4-2h^2r^2+r^4-8h^2-8r^2+16=0$

Then solve this quartic in $c^2=(\frac1{b})^2$ for given $h,k,r.$

Old first attempt

The dual curves are $4X^2+b^2Y^2=1$ and $10XY+24Y^2+2X+10Y+1=0.$ The common tangents are given by $x(P_i)x+y(P_i)y+1=0,$ where $P_i$ are the intersection points of the duals. When $b\approx 4.26941996$ the inner common tangents coincide, as we can see using a slider in geogebra.

Old secoond attempt

It may seem we reduced to the same type of situation, finding when the dual curves are tangent. But, now $b$ does not appear in a fraction.

So, now we can find the tact invariant as in 3264 conics in a second in M2

R=QQ[x,y,b,MonomialOrder=>Eliminate 2]
I=ideal(10*x*y+24*y^2+2*x+10*y+1,-4*x^2-b^2*y^2+1,determinant matrix{{diff(x,10*x*y+24*y^2+2*x+10*y+1),diff(x,-4*x^2-b^2*y^2+1)},{diff(y,10*x*y+24*y^2+2*x+10*y+1),diff(y,-4*x^2-b^2*y^2+1)}}) -- b^6-81*b^4+2187*b^2-19008

giving a cubic in $b^2.$ The roots in maxima CAS:

allroots(b^6-81*b^4+2187*b^2-19008);
  [b = 4.269419958889195,b = -4.269419958889191,
   b = 0.6731641510895302*%i-5.642621428168951,
   b = (-0.6731641510895302*%i)-5.642621428168951,
   b = 0.6731641510895462*%i+5.642621428168949,
   b = 5.642621428168949-0.6731641510895462*%i]