# Find the radius of the circle tangent to $x^2, \sqrt{x}, x=2$

I'm taking up a post that was closed for lack of context because I'm very interested in it :

Let $$(a,b)$$ be the center of this circle. It seems intuitive that $$b=a$$, but I have not been able to prove it formally, although I know that two reciprocal functions are symmetric with respect to the first bisector $$y=x$$.

Then let $$(X,X^2)$$ the point of tangency with $$y=x^2$$. I think we're going to use the formula for the distance from $$(a,a)$$ to the line $$y-X^2=2X(x-X)$$.

We have obviously the relation $$r=2-a$$. The normal to $$(X,X^2)$$ passes through $$(a,a)$$

I'm not sure if my notations are the best to elegantly solve the exercise. I hope you will share my enthusiasm for this lovely exercise that I have just discovered thanks to MSE.

• Note that the tangent lines have slopes that are reciprocals so yes, the center must be on the line $y=x$ Commented Mar 21 at 20:17
• Well, $r=2-\frac{2x^3+x}{2x+1}$, where $x$ is the root of $8x^5-17x^4-6x^3+8x^2+12x+4=0$ in $(1,2)$. So $r\approx 0.344556$. $x$ is $X$ in your notation. $x \approx 1.33284$. I failed to solve this equation exactly though.
– Aig
Commented Mar 21 at 22:01
• @Aig : yes , i obtain the same . Maybe we can only obtain an approximation. Commented Mar 21 at 22:03
• Maybe it might be clearest just to write out the equations relating $x_0, y_0, r, x_1, y_1, x_2, y_2, x_3, y_3$ where the circle is defined by $(x_0, y_0), r$ and $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ are the tangency points. So for example, $y_1 = x_1^2$, $\frac{y_1 - y_0}{x_1 - x_0} \cdot (2x_1) = -1$, $(y_1-y_0)^2+(x_1-x_0)^2=r^2$ and convert the second into a polynomial equation. Then, put those into some Groebner basis calculator (using a monomial order to eliminate in favor of $x_0,y_0,r$) and let that take care of things... Commented Mar 21 at 22:59
• SageMath reports the Galois group of the splitting field of $8x^5-17x^4-6x^3+8x^2+12x+4$ over $\mathbb{Q}$ is $S_5$ so the polynomial is not solvable by radicals. Commented Mar 21 at 23:07

From symmetry, it can be assumed at first that the center of the tangent circle is on the line $$y = x$$.

So, let the center of the circle be at $$(a,a)$$, and let the tangency point with the parabola $$y = x^2$$ be $$(b , b^2)$$, then the radius of the circle is

$$r = | 2 - a |$$

The vector $$(a - b , a - b^2)$$ is along the normal vector to the parabola, which given by its gradient vector, namely $$(2 x, -1)$$. Therefore,

$$(a - b , a - b^2) = K (2 b , -1)$$

So that

$$(-1) ( a - b) - 2 b (a - b^2) = 0$$

And finally, the radius is equal to the distance between the the two points $$(a, a)$$ and $$(b , b^2)$$, hence

$$(a - 2)^2 = (a - b)^2 + (b^2 - a)^2$$

The solutions of these equations are

$$(a, b) = (1.655444, 1.332841)$$ and $$(3.783181, 2.050749)$$ and $$(-5.21545, -0.44123)$$

The corresponding radii are $$0.344556$$ , $$1.783181$$, and $$7.21545$$ respectively. The two circles corresponding to the first two of these solutions are shown below.

Now, if we want all the possible solutions of circles tangent to the two parabolas and the vertical line $$x = 2$$, then we'll take the center of the circle to be

$$C = (a, b)$$

And the tangency point with $$y = x^2$$ to be

$$r_1 = (c, c^2)$$

And the tangency point with $$x = y^2$$ to be

$$r_2 = (d^2, d )$$

For convenience define

$$e = c^2$$ and $$f = d^2$$

We now want the vector $$(C - r_1)$$ to be along the normal vector to $$y = x^2$$ at $$r_1$$, i.e.

$$C - r_1 = K_1 ( 2 c, - 1)$$

And similarly, we want $$(C - r_2)$$ to be along the normal vector to $$x = y^2$$ at $$r_2$$, i.e.

$$C - r_2 = K_2 ( 1, - 2 d )$$

Moreover, we have the radius of the circle equal to $$| a - 2 |$$ , so we want

$$| a - 2 |^2 = \| C - r_1 \|^2$$

and

$$| a - 2 |^2 = \| C - r_2 \|^2$$

In terms of our $$6$$ unknowns $$a,b,c,d,e,f$$, we have the following $$6$$ equations relating them

$$e = c^2$$

$$f = d^2$$

$$- (a - c) - 2 c (b - e) = 0$$

$$- 2 d ( a - f) - (a - d) = 0$$

$$(a - 2 )^2 = (a - c)^2 + (b - e)^2$$

$$(a - 2)^2 = (a - f)^2 + (b - d)^2$$

This is a quadratic system of $$6$$ equations in $$6$$ unknowns. It can be solved via iterative methods such the Newton-Raphson multivariate method. I used this method with equally spaced initial guess for $$a$$ and $$b$$ , from which I computed the initial guess for $$c$$ and $$d$$ and from that, the initial guess for $$e$$ and $$f$$ follows directly. So changing $$(a,b)$$ over a grid centered at the origin, with step size equal to $$0.5$$ in each direction, I obtained $$8$$ solutions, out of which only $$5$$ of them have their $$d \gt 0$$. These are the ones retained because we want tangency with $$y = \sqrt{x} \gt 0$$. These $$5$$ circles have their center $$(a,b)$$ and radius $$r$$ as follows:

$$\begin{array}{|c|c|c|} \hline \\ a && b && r \\ \hline \\ 1.422724 && 0.562507 && 0.577276\\ -4.158013 && 4.800189&& 6.158013\\ 0.242594 && 2.58606 && 1.757406\\ 3.783181 && 3.783181 && 1.783181 \\ 1.655444&& 1.655444&&0.344556 \\ \hline \end{array}$$

These $$5$$ circle are shown in the figure below.

• The argument "From symmetry" is not clear. See @Piquito's answer. Commented Mar 22 at 7:42
• What do you mean it is not clear ? Symmetry is very clear in the problem. Commented Mar 22 at 7:45
• (1) What did you use to solve the equations? The accuracy of your solutions is quite low. Wolfram Alpha gives $a\approx1.65544$, $b\approx1.33284$ (this agrees with my comment to the post). (2) The symmetry argument does not look clear at all.
– Aig
Commented Mar 22 at 8:45
• @Aig I've used my own numerical method to find the solutions, and you're right the accuracy for some reason unknown to me is not very good in this particular system of equations. I've check the solution of the system in SAGE and got the results that you reported as well. (2) As for symmetry, if you place a circle such that it is tangent to both $y = x^2$ and $x = y^2$ , then I think it is obvious that the center of the circle must lie on $y = x$. Commented Mar 22 at 10:20
• @Aig Turns out that I made an error in translating one of the two equations into code. Now that error is fixed, and I got the same solutions as those reported by Wolframalpha. Commented Mar 23 at 14:25

I let $$(t,t^2)$$ be one tangent point $$(s^2,s)$$ another and $$(2,y_2)$$ the third. The first three equations then expresses these points are on the circle, the two next compare the equations with scaling factor l2 of the tangents to the circle at $$(t,t^2)$$ and the parabola $$y=x^2$$ at the same point. The scaling factor l3 the same for $$(s^2,s)$$ on $$x=y^2.$$ the last equation says that y2$$=k$$ or that the $$y$$-coordinate of the last tangent point is the same as that of the center, essentially that this is a tangent point for the circle and $$x=2.$$

Then taking the grobner basis in M2 with an elimination order and keeping only elements with $$h,k,r$$ and feeding them to maxima CAS solve I get the eight real solutions pictured of which five are tangent to the positive branch of $$\sqrt{x}$$. I also get these eight repeated with negative $$r.$$ And lots of non-real solutions (40).

R=QQ[l2,l3,s,t,y2,h,k,r,MonomialOrder=>Eliminate 5]
S=R[x,y]
(x+2-h)^2+y^2-r^2
toString oo
(x+s^2-h)^2+(y+s-k)^2-r^2
toString oo
(x+t-h)^2+(y+t^2-k)^2-r^2
toString oo


The first three equations are the zero sets of the constant terms wrt the total order on $$x,y$$ in the above three. Then the linear terms are the tengent( cone)s.

I=ideal(h^2-r^2-4*h+4,s^4-2*s^2*h+s^2-2*s*k+h^2+k^2-r^2,t^4-2*t^2*k+t^2-2*t*h+h^2+k^2-r^2)


The above line encodes the first three eqautions.

(x-s^2)-2*s*(y-s)-l2*((s^2-h)*x+(s-k)*y-s^4+s^2*h-s^2+s*k)
toString oo
-2*t*x+y+t^2-l3*((2*t-2*h)*(x-t)+(2*t^2-2*k)*(y-t^2)+t^4-2*t^2*k+t^2-2*t*h+h^2+k^2-r^2)
toString oo


The next two are the zero sets of the coefficients wrt $$x,y,1$$ in the above. These are the equations of the tangents of the circle and parabolas equated with a scaling factor, which must vanish.

use R
J=I+ideal((-l2*s^2+l2*h+1),(-l2*s+l2*k-2*s),l2*s^4-l2*s^2*h+l2*s^2-l2*s*k+s^2)+ideal((-2*l3*t+2*l3*h-2*t),(-2*l3*t^2+2*l3*k+1),l3*t^4+l3*t^2+t^2-l3*h^2-l3*k^2+l3*r^2)+ideal(y2-k)
gens gb J


In the above grobner basis, produced by the last line, pick out the h,k,r terms and feed it to maxima CAS

solve([h^2-r^2-4*h+4,64*k^3*r^2+128*h*k^3-16*k^4+288*h*k*r^2-280*k^2*r^2+27*r^4-640*h*k^2-120*k^3-952*h*r^2+346*k*r^2+1288*h*k+895*k^2-1111*r^2-3104*h-1928*k+5168,256*k^6+256*k^2*r^4+1408*h*k^4-128*k^5-128*h*k^2*r^2+1024*h*r^4+216*k*r^4-640*h*k^3-2528*k^4+2464*h*k*r^2-6840*k^2*r^2-2127*r^4-16224*h*k^2+2960*k^3-37928*h*r^2+3222*k*r^2+20248*h*k+22045*k^2-28645*r^2-101920*h-26136*k+170576, 131072*k^2*r^6+196608*h*k^2*r^4+524288*h*r^6+110592*k*r^6+16384*h*k^5+1692672*h*k*r^4-5239296*k^2*r^4-3296*r^6+30720*h*k^4-18304*k^5-18096896*h*k^2*r^2-20312512*h*r^4+5500904*k*r^4-313728*h*k^3+229680*k^4+17765472*h*k*r^2+44536312*k^2*r^2-57943053*r^4+53469568*h*k^2-8229936*k^3-53207736*h*r^2-28873310*k*r^2-50714808*h*k-71874089*k^2+330702801*r^2+322842336*h+54506424*k-526193744,2097152*h*k*r^6+1769472*r^8-48857088*h*k^2*r^4+24281088*h*r^6+32254976*k*r^6+2670592*h*k^5+272188416*h*k*r^4-317489920*k^2*r^4-103776544*r^6+2025472*h*k^4-2567808*k^5-1064738048*h*k^2*r^2-1298567872*h*r^4+638613688*k*r^4-45057152*h*k^3+11526160*k^4+2583167776*h*k*r^2+2138573096*k^2*r^2-2235903423*r^4+1875356416*h*k^2-275371728*k^3-3094227176*h*r^2-2642803562*k*r^2-585489512*h*k-2471857843*k^2+14236910699*r^2+12228329888*h-301365912*k-19894511216,226492416*k*r^8-3504996352*r^8+77798440960*h*k^2*r^4-49916084224*h*r^6-36070443008*k*r^6-4125442048*h*k^5-340115567616*h*k*r^4+527530298368*k^2*r^4+106054310240*r^6-3306739712*h*k^4+3927668096*k^5+1782003758848*h*k^2*r^2+1766756044736*h*r^4-983060549512*k*r^4+70285139840*h*k^3-19552394992*k^4-3770245003744*h*k*r^2-3598165946456*k^2*r^2+3479512727153*r^4-3212301699200*h*k^2+478661522288*k^3+3305646410520*h*r^2+3948768222326*k*r^2+1459051956248*h*k+4231170848317*k^2-22195490619397*r^2-21807930807392*h-276838006552*k+35631383791760,97844723712*h*r^8+1240938119168*r^8-11784687878144*h*k^2*r^4+9974603743232*h*r^6+768346485760*k*r^6-258675490816*h*k^5+12851705502720*h*k*r^4-86013973955840*k^2*r^4-20331080028640*r^6+832387569664*h*k^4+257740337792*k^5-265709649730304*h*k^2*r^2-342359069108032*h*r^4+17609058671240*k*r^4-13299724339072*h*k^3+2004580875632*k^4-26136867024928*h*k*r^2+806348640917656*k^2*r^2-473713110069865*r^4+1005682425369856*h*k^2-118990257993904*k^3+604175783904552*h*r^2-381668227612678*k*r^2-1271031404594008*h*k-1398258797425493*k^2+4318862224361693*r^2+6325196568001888*h+1647569763185240*k-10478733245626384, 285315214344192*r^10-34313452342476800*r^8+288757288168128512*h*k^2*r^4-316658753199276032*h*r^6-5366998348496896*k*r^6+13269104141418496*h*k^5-270957021877103616*h*k*r^4+2513026818947113472*k^2*r^4-15614238496397792*r^6-22960130057193472*h*k^4-13303434366567296*k^5+8578387111462491392*h*k^2*r^2+7341226741309417024*h*r^4-337540265138525912*k*r^4+186583391288322688*h*k^3-63600665880951632*k^4+1864930142528362336*h*k*r^2-23980536595923508552*k^2*r^2+16065097321181450923*r^4-29651105263966235776*h*k^2+4013975526908095312*k^3-20835543130198795128*h*r^2+10643968418844203122*k*r^2+39853365467734514824*h*k+40636356901971413807*k^2-128053996524505779527*r^2-194790094752513635872*h-51751948576206932360*k+322404294920479922992],[h,k,r]);


Maxima spits out the following in a second or two.

[[h = -4.158012980529206,k = 4.800188501413761,r = 6.158013544018059],
[h = 0.2196604233287924-4.227035900570452*%i,
k = (-2.873309177540616*%i)-3.127055638404707,
r = 4.227035900570609*%i+1.780339576671407],
[h = 4.227035900570452*%i+0.2196604233287924,
k = 2.873309177540615*%i-3.127055638404706,
r = 1.780339576671207-4.227035900570453*%i],
[h = 8.018555334658715,k = 3.28691904047976,r = -6.018555334658714],
[h = 0.7962439166561859-3.592036258951989*%i,
k = 2.376465459305944*%i+2.897149125584387,
r = 3.592036258951982*%i+1.203756083343819],
[h = 3.592036258951989*%i+0.7962439166561859,
k = 2.897149125584387-2.376465459305945*%i,
r = 1.203756083343813-3.592036258951989*%i],
[h = (-1.814443734727141*%i)-2.173025101865336,
k = 0.2397075303832508*%i+4.52699497605145,
r = 1.814443734727141*%i+4.173025101865336],
[h = 1.814443734727141*%i-2.173025101865336,
k = 4.526994976051534-0.2397075303831821*%i,
r = 4.173025101865333-1.814443734727127*%i],
[h = 0.2360429802241225-3.111501495423057*%i,
k = 2.009074994746165*%i+2.909197802035789,
r = 3.111501495423057*%i+1.763957019775876],
[h = 3.111501495423057*%i+0.2360429802241225,
k = 2.90919780203577-2.00907499474618*%i,
r = 1.763957019775873-3.111501495423075*%i],
[h = 1.422723694568354,k = 0.5625075057043353,r = 0.5772763054316452],
[h = 1.0549380030495-0.7427611205817737*%i,
k = (-0.4946483114830123*%i)-2.077179855544079,
r = 0.7427611205817709*%i+0.9450619969505265],
[h = 0.7427611205817737*%i+1.0549380030495,
k = 0.494648311483019*%i-2.077179855544091,
r = 0.9450619969505257-0.7427611205817711*%i],
[h = 1.166866535526935-0.4878771706929158*%i,
k = (-0.8976802483452873*%i)-0.7828852096796389,
r = 0.4878771706929925*%i+0.8331334644731291],
[h = 0.4878771706929158*%i+1.166866535526935,
k = 0.8976802483452877*%i-0.7828852096796393,
r = 0.8331334644731291-0.4878771706929924*%i],
[h = 1.081700367406306-0.3054769894307804*%i,
k = (-0.6547395459409551*%i)-0.6083327611478294,
r = 0.3054769894308998*%i+0.9182996325939112],
[h = 0.3054769894307804*%i+1.081700367406306,
k = 0.6547395459409554*%i-0.6083327611478293,
r = 0.9182996325939112-0.3054769894308998*%i],
[h = 1.023851661320742-0.2556031361188028*%i,
k = (-0.4648234474984263*%i)-0.06042095744029741,
r = 0.2556031361188198*%i+0.9761483386794274],
[h = 0.2556031361188028*%i+1.023851661320742,
k = 0.4648234474984263*%i-0.06042095744029743,
r = 0.9761483386794316-0.2556031361188153*%i],
[h = 1.192745676929565,k = -0.1891523111035306,r = 0.8072543230704344],
[h = 0.6013058651242753-0.2484954256656225*%i,
k = 0.6446370182184701*%i+0.7243289206238143,
r = 0.2484954256656233*%i+1.398694134875725],
[h = 0.2484954256656225*%i+0.6013058651242753,
k = 0.7243289206238143-0.6446370182184702*%i,
r = 1.398694134875725-0.2484954256656233*%i],
[h = 0.2425937435259995,k = 2.586060122372971,r = 1.757406372275014],
[h = -5.21544885177453,k = -5.21544885177453,r = 7.21544885177453],
[h = 3.783180778032037,k = 3.783180778032037,r = -1.783180778032036],
[h = 1.655444126074498,k = 1.655444126074498,r = 0.3445559845559846],
[h = 0.7399744850693703-0.07364991216678059*%i,
k = 0.7399744850693566-0.07364991216676495*%i,
r = 0.07364991216678053*%i+1.260025514930629],
[h = 0.07364991216678059*%i+0.7399744850693703,
k = 0.07364991216676505*%i+0.7399744850693567,
r = 1.260025514930629-0.07364991216678056*%i],
[h = -4.158012980529206,k = 4.800188501413761,r = -6.158013544018059],
[h = 0.2196604233287924-4.227035900570452*%i,
k = (-2.873309177540616*%i)-3.127055638404707,
r = (-4.227035900570451*%i)-1.780339576671209],
[h = 4.227035900570452*%i+0.2196604233287924,
k = 2.873309177540615*%i-3.127055638404706,
r = 4.227035900570494*%i-1.780339576671212],
[h = 8.018555334658715,k = 3.28691904047976,r = 6.018555334658714],
[h = 0.7962439166561859-3.592036258951989*%i,
k = 2.376465459305944*%i+2.897149125584387,
r = (-3.592036258951989*%i)-1.203756083343813],
[h = 3.592036258951989*%i+0.7962439166561859,
k = 2.897149125584387-2.376465459305945*%i,
r = 3.592036258951983*%i-1.20375608334379],
[h = (-1.814443734727141*%i)-2.173025101865336,
k = 0.2397075303832508*%i+4.52699497605145,
r = (-1.814443734727141*%i)-4.173025101865336],
[h = 1.814443734727141*%i-2.173025101865336,
k = 4.526994976051534-0.2397075303831821*%i,
r = 1.814443734727123*%i-4.173025101865336],
[h = 0.2360429802241225-3.111501495423057*%i,
k = 2.009074994746165*%i+2.909197802035789,
r = (-3.111501495423057*%i)-1.763957019775876],
[h = 3.111501495423057*%i+0.2360429802241225,
k = 2.90919780203577-2.00907499474618*%i,
r = 3.111501495423078*%i-1.763957019775875],
[h = 1.422723694568354,k = 0.5625075057043353,r = -0.5772763054316452],
[h = 1.0549380030495-0.7427611205817737*%i,
k = (-0.4946483114830123*%i)-2.077179855544079,
r = (-0.7427611205817712*%i)-0.9450619969505262],
[h = 0.7427611205817737*%i+1.0549380030495,
k = 0.494648311483019*%i-2.077179855544091,
r = 0.7427611205817712*%i-0.9450619969505257],
[h = 1.166866535526935-0.4878771706929158*%i,
k = (-0.8976802483452873*%i)-0.7828852096796389,
r = (-0.4878771706929926*%i)-0.8331334644731292],
[h = 0.4878771706929158*%i+1.166866535526935,
k = 0.8976802483452877*%i-0.7828852096796393,
r = 0.4878771706929916*%i-0.8331334644731319],
[h = 1.081700367406306-0.3054769894307804*%i,
k = (-0.6547395459409551*%i)-0.6083327611478294,
r = (-0.3054769894308997*%i)-0.9182996325939112],
[h = 0.3054769894307804*%i+1.081700367406306,
k = 0.6547395459409554*%i-0.6083327611478293,
r = 0.3054769894308997*%i-0.9182996325939112],
[h = 1.023851661320742-0.2556031361188028*%i,
k = (-0.4648234474984263*%i)-0.06042095744029741,
r = (-0.2556031361188198*%i)-0.9761483386794273],
[h = 0.2556031361188028*%i+1.023851661320742,
k = 0.4648234474984263*%i-0.06042095744029743,
r = 0.2556031361188153*%i-0.9761483386794316],
[h = 1.192745676929565,k = -0.1891523111035306,
r = -0.8072543230704344],
[h = 0.6013058651242753-0.2484954256656225*%i,
k = 0.6446370182184701*%i+0.7243289206238143,
r = (-0.2484954256656235*%i)-1.398694134875725],
[h = 0.2484954256656225*%i+0.6013058651242753,
k = 0.7243289206238143-0.6446370182184702*%i,
r = 0.2484954256656234*%i-1.398694134875725],
[h = 0.2425937435259995,k = 2.586060122372971,r = -1.757406372275014],
[h = -5.21544885177453,k = -5.21544885177453,r = -7.21544885177453],
[h = 3.783180778032037,k = 3.783180778032037,r = 1.783180778032036],
[h = 1.655444126074498,k = 1.655444126074498,r = -0.3445559845559846],
[h = 0.7399744850693703-0.07364991216678059*%i,
k = 0.7399744850693566-0.07364991216676495*%i,
r = (-0.07364991216678058*%i)-1.260025514930629],
[h = 0.07364991216678059*%i+0.7399744850693703,
k = 0.07364991216676505*%i+0.7399744850693567,
r = 0.07364991216678055*%i-1.260025514930629]]


The rest was taking the eight real solutions with positive $$r$$ in to geogebra.

PS: The lex $$h,k,r$$ grobner basis has three elements, a first element depending only on r of degree $$56$$ that factors as two degree five (that encodes the three real symmetric solutions as three and three with both the positive and negative r solutions) and two degree $$23$$ polynomials (that take care of the five other real solutions as five and five with both the positive and negative r solutions).

(64*r^5-531*r^4+330*r^3+1468*r^2-1836*r+452)*(64*r^5+531*r^4+330*r^3-1468*r^2-1836*r-452)*(12230590464*r^23-382404395008*r^22+5674381471744*r^21-51704974992576*r^20+299776276321152*r^19-864625483662144*r^18-2700361220009408*r^17+51202787746801056*r^16-363287294911724512*r^15+1772609715198384952*r^14-6635617359623812536*r^13+19805852496404944821*r^12-47885587742865828330*r^11+94327515582292632476*r^10-151472653220772515888*r^9+197705207462366063666*r^8-208464366358887460352*r^7+175816898388784166704*r^6-116813669362464313276*r^5+59738270712662252581*r^4-22675656056778017622*r^3+6013321025916623244*r^2-993684774850618620*r+76955803214628100)*(12230590464*r^23+382404395008*r^22+5674381471744*r^21+51704974992576*r^20+299776276321152*r^19+864625483662144*r^18-2700361220009408*r^17-51202787746801056*r^16-363287294911724512*r^15-1772609715198384952*r^14-6635617359623812536*r^13-19805852496404944821*r^12-47885587742865828330*r^11-94327515582292632476*r^10-151472653220772515888*r^9-197705207462366063666*r^8-208464366358887460352*r^7-175816898388784166704*r^6-116813669362464313276*r^5-59738270712662252581*r^4-22675656056778017622*r^3-6013321025916623244*r^2-993684774850618620*r-76955803214628100)


The last two relate $$h, k$$ linearly to degree $$54$$ polynomials in these $$r.$$

• You're talking about "The first three equations". What "first three equations"? Commented Mar 25 at 7:44
• @StéphaneJaouen See the edit Commented Mar 25 at 10:23
• Thanks, but I don't know this computer program or what a grobner basis is. Don't you think we can propose a more basic solution? Commented Mar 25 at 10:31
• @StéphaneJaouen That requires a bit more thought. I'll see what I can do Commented Mar 25 at 10:48

Answer. $$r\approx 0.344555975$$. More precisely, $$r$$ is the medium real root of the polynomial $$64r^5-531r^4+330r^3+1468r^2-1836r+452.$$

Solution. Let us justify the symmetry arguments first. Namely, we claim that the center $$(x_0,y_0)$$ of the circle belongs to the bisector $$x=y$$, even when we relax the condition that the circle is tangent to the line $$x=2$$. Indeed, let the circle is tangent to the graph $$y=x^2$$ at a point $$(x,x^2)$$. Then the vector $$(x-x_0,x^2-y_0)$$ has length $$r$$ and is orthogonal to the tangent vector $$(1,2x)$$ to the graph $$y=x^2$$ at the point $$(x,x^2)$$. It follows $$x-x_0=\frac{-2rx}{\sqrt{1+4x^2}}$$ and $$x^2-y_0=\frac{r}{\sqrt{1+4x^2}}$$. Similarly, if the circle is tangent to the graph $$x=y^2$$ at a point $$(y^2,y)$$ then $$y-y_0=\frac{-2ry}{\sqrt{1+4y^2}}$$ and $$y^2-x_0=\frac{r}{\sqrt{1+4y^2}}$$. Thus $$x_0=x+\frac{2rx}{\sqrt{1+4x^2}}=y^2-\frac{r}{\sqrt{1+4y^2}}$$ and $$y_0=x^2-\frac{r}{\sqrt{1+4x^2}}=y+\frac{2ry}{\sqrt{1+4y^2}}.$$ So $$x_0+y_0=f(x)=f(y)$$, where $$f(t)=t^2+t+\frac{r(2t-1)}{\sqrt{1+4t^2}}$$. Since $$f'(t)=(2t+1)\left(1+\frac{2r}{(1+4t^2)^{3/2}}\right)>0$$, the function $$f$$ is increasing. Since $$f(x)=f(y)$$, we have $$x=y$$, and so $$x_0=y_0$$. Thus $$x+\frac{2rx}{\sqrt{1+4x^2}}=x^2-\frac{r}{\sqrt{1+4x^2}}$$, $$r=\frac{(x^2-x)\sqrt{1+4x^2}}{1+2x}$$, $$x_0=\frac{2x^3+x}{1+2x}$$.

So when we fix $$x_0+r=2$$ then we obtain $$\frac{2x^3+x}{1+2x}+\frac{(x^2-x)\sqrt{1+4x^2}}{1+2x}=2,$$ which implies $$8x^5-17x^4-6x^3+8x^2+12x+4=0.$$

We already met this equation. According to Mathcad calculations, it has three real roots: $$-.441229\dots$$, $$1.332840\dots$$, and $$2.050748\dots$$. Since $$0, only the second root is feasible, and it provides $$r= 0.34455597499666\dots.$$

Now let us build a polynomial with a root $$r$$. We have $$r=\frac{(x^2-x)\sqrt{1+4x^2}}{1+2x}=2-\frac{2x^3+x}{1+2x}.$$ Thus $$2x^3+(2r-3)x+r-2=0$$. Since the polynomials $$8x^5-17x^4-6x^3+8x^2+12x+4$$ and $$2x^3+(2r-3)x+r-2=0$$ have a common root $$x$$, their resultant

$$\left|\begin{matrix} 8 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ -17 & 8 & 0 & 0 & 2 & 0 & 0 & 0 \\ -6 & -17 & 8 & 2r-3 & 0 & 2 & 0 & 0 \\ 8 & -6 & -17 & r-2 & 2r-3 & 0 & 2 & 0 \\ 12 & 8 & -6 & 0 & r-2 & 2r-3 & 0 & 2 \\ 4 & 12 & 8 & 0 & 0 & r-2 & 2r-3 & 0 \\ 0 & 4 & 12 & 0 & 0 & 0 & r-2 & 2r-3 \\ 0 & 0 & 4 & 0 & 0 & 0 & 0 & r-2 \end{matrix}\right|=$$ $$18(64r^5-531r^4+330r^3+1468r^2-1836r+452)$$

equals zero. According to Mathcad calculations, it the resultant has, besides $$r$$, two reals roots: $$-1.783180\dots$$ and $$7.215448\dots$$. Unfortunately, as I understood Daniel Schepler's comment it implies that the respective quintic equation cannot be solved in radicals.

Just to avoid numerical methods.

Cheating a little, the solution of $$8x^5-17x^4-6x^3+8x^2+12x+4=0$$ is close to $$\frac 43$$. Let $$x=\left(\frac 43+t\right)$$ to face $$f(t)=1944 t^5+8829 t^4+11070 t^3-1872 t^2-8124 t-4$$ Now, rewrite (this is tricky) $$f(t)=-4-8124 t-1872 t^2+11070 t^3+8829 t^4+1944 t^5+O\left(t^{p}\right)$$ and use power series reversion.

If $$p=6$$, this gives as an estimate $$t_{(6)}=-z-\frac{156 }{677}z^2-\frac{1346409 }{916658}z^3-\frac{675607473 }{1241154932}z^4-\frac{4928871973923 }{840261888964}z^5+O\left(z^6\right)$$ where $$z=\frac{f(t)+4}{8124}$$ Since we want $$f(t)=0$$, $$z=\frac{1}{2031}$$ and then $$t_{(6)}=-\frac{44132524856806035483893}{89622957837131547712586511}\implies x_{(6)}=\frac{39817714863772863638210485}{29874319279043849237528837}$$ which is in an absolute error of $$2.60\times 10^{-20}$$.

Continuing $$x_{(7)}=\frac{145996907486385222544671695804265}{109538135046759107017498 834666984}$$ is in an absolute error of $$2.17\times 10^{-22}$$.

$$x_{(8)}=\frac{66914616611327452663665908822936673079}{5020450389784605476 0223223394084109736}$$ is in an absolute error of $$2.02\times 10^{-26}$$.

We could go as far as we want using the explicit formula for the $$n^{\text{th}}$$ term as given by Morse and Feshbach.

• how does writing everything in fractions "avoid numerical methods"? Commented Mar 26 at 11:41

1.- Regarding the circle proposed by @Piquito, working with geogebra, I come to the conclusion that such a circle does indeed exist. It seems that its center is on the line $$y=2-x$$ and that it is tangent to $$y=0$$, which makes the $$x=2$$ line very interesting. https://www.geogebra.org/m/hcchm2tx

2.- By the way, I think it's a shame not to show the curves updated by @Aig in one of his links in the comments :

The first is obtained by writing:$$\begin{vmatrix}a-b & 2b \\a-b^2 & -1\end{vmatrix}=0$$

We have the parabolas $$\mathscr{P}_1,\mathscr{P}_2$$ and line $$\mathscr{L}$$

$$\cases{ \mathscr{P}_1\to y-x^2=0\\ \mathscr{P}_2\to y^2-x=0\\ \mathscr{L}\to x - 2 = 0 }$$

and we are interested in drawing the circles simultaneously tangent to these three elements. The proposed solution process consists of immersing the particular problem in a family of problems that incorporate it.

Let the generic circle be $$\mathscr{C}\to (x-x_0)^2+(y-y_0)^2-r^2=0$$. Now, if $$\mathscr{P}_1$$ and $$\mathscr{C}$$ are tangent, then eliminating $$x$$ between them we arrive at the polynomial

$$p_1(y) = r^4 - 2 r^2 x_0^2 + x_0^4 - 2 r^2 y - 2 x_0^2 y + y^2 - 2 r^2 y^2 + 2 x_0^2 y^2 + 2 y^3 + y^4 + 4 r^2 y y_0 - 4 x_0^2 y y_0 - 4 y^2 y_0 - 4 y^3 y_0 - 2 r^2 y_0^2 + 2 x_0^2 y_0^2 + 2 y y_0^2 + 6 y^2 y_0^2 - 4 y y_0^3 + y_0^4 = 0$$

This polynomial should have a double root in $$y$$ due to tangency hence

$$\mathscr{P}_1\cap \mathscr{C}\to p_1(y) \equiv c_0(y-y_1)^2(y^2+c_1 y + c_2)\ \ \ \forall y\ \ \ \ \ (1)$$

and analogously

$$\cases{ \mathscr{P}_1\cap \mathscr{C}\to p_1(y) \equiv c_0(y-y_1)^2(y^2+c_1 y + c_2)\\ \mathscr{P}_2\cap \mathscr{C}\to p_2(y) \equiv d_0(y-y_2)^2(y^2+d_1 y + d_2)\\ \mathscr{L}\cap \mathscr{C}\to p_3(y) \equiv e_0(y-y_2)^2\\ }$$

Here $$c_i,d_i,e_i,y_1,y_2,y_2$$ are coefficients to determine. Note that $$p_3(y)$$ has maximum degree $$2$$. Now from $$(1)$$ we have

$$\cases{ r^4 - 2 r^2 x_0^2 + x_0^4 - 2 r^2 y_0^2 + 2 x_0^2 y_0^2 + y_0^4 - c_0 c_2 y_1^2 = 0\\ 4 r^2 y_0 -2 r^2 - 2 x_0^2 - 4 x_0^2 y_0 + 2 y_0^2 - 4 y_0^3 + 2 c_0 c_2 y_1 - c_0 c_1 y_1^2 = 0\\ 1 - c_0 c_2 - 2 r^2 + 2 x_0^2 - 4 y_0 + 6 y_0^2 + 2 c_0 c_1 y_1 - c_0 y_1^2=0\\ 2 - c_0 c_1 - 4 y_0 + 2 c_0 y_1 = 0\\ 1 - c_0 = 0 }$$

and thus we can proceed for the remaining $$p_i(y)$$ obtaining a set of $$13$$ conditions in $$13$$ unknowns to solve.

Follows a MATHEMATICA script which handles the symbolic tasks and solves the resulting system of conditions. In the script, some trivial coefficients as $$c_0,d_0,e_0$$ are explicitly assumed.

cond0 = First[Eliminate[{x - 2 == 0, (x - x0)^2 + (y - y0)^2 - r^2 == 0}, x]] + (y - y00)^2;
cond1 = First[Eliminate[{y - x^2 == 0, (x - x0)^2 + (y - y0)^2 - r^2 == 0},x]] - (y - y01)^2 (y^2 + c1 y + c2);
cond2 = First[Eliminate[{y^2 - x == 0, (x - x0)^2 + (y - y0)^2 - r^2 == 0},x]] - (y - y02)^2 (y^2 + c3 y + c4);

coefs0 = CoefficientList[cond0, y];
coefs1 = CoefficientList[cond1, y];
coefs2 = CoefficientList[cond2, y];

coefs = Join[Join[coefs0, coefs1], coefs2];
vars = {y00, y01, y02, c1, c2, c3, c4, r, x0, y0};
sols = Solve[coefs == 0, vars, Reals] // N;
circs = {x0, y0, r} /. sols // Chop;

feas = {};
lim = 200;
For[k = 1, k <= Length[circs], k++,
If[Im[circs[[k, 3]]] == 0 && lim > circs[[k, 3]] > 0,
AppendTo[feas, circs[[k]]]]
]

gr0 = Table[Graphics[{Dashed, Circle[Take[feas[[k]], {1, 2}], Abs[feas[[k, 3]]]]}], {k, 1, Length[feas]}];
gr1 = ContourPlot[{x^2 == y, x == y^2, x == 2}, {x, -10, 10}, {y, -10, 10}];
Show[gr0, gr1, PlotRange -> {{-3, 6}, {-2, 7}}, AspectRatio -> 1]


COMMENT.-Your idea of using the normal to a curve is good.You do have two solutions given by the two circles $$\left(x-\dfrac53\right)^2+\left(y-\dfrac53\right)^2=\left(\dfrac13\right)^2\\(x-1.43)^2+(y-0.57)^2=0.56^2$$

Your problem is nice and I leave you to solve it by yourself.

• Well, the lower circle certainly does not have its center on the line $y=x$. So the symmetry argument in the OP is suspect, I guess. I haven't thought about it yet. Commented Mar 21 at 21:23
• What about a circle to the right of $x=2$? Commented Mar 21 at 21:25
• Are you sure about the first circle's parameters? Something seems a bit off Commented Mar 21 at 21:29
• @piquito There are more than two solutions. Commented Mar 21 at 21:41
• @user376343: Yes I see it now. Thanks. Commented Mar 22 at 22:50

Some solutions are on the line $$y = x$$. The reverse is not true as shown by Jan-Magnus Økland.

The reason is that $$y = \sqrt{x}$$ is the inverse function of $$y = x^2$$ for $$x\geq 0$$ hence its plot is the reflection of the plot of $$y = x^2$$ over the line $$y = x$$. Hence for any circle on the line $$y = x$$ that touches $$y = x^2$$, it also touches $$y = \sqrt{x}$$. The line $$x = 2$$ then can be made tangent by moving the center of the circle on the line $$y = x$$.

Let $$T(x_t,x^2_t)$$ be the tangent point between the circle to the parabola. The tangent vector at T is then $$(1,2x)$$, consequently: $$(a-x_t) + (a-x^2_t)2x_t=0 \Rightarrow a = x_t\frac{1+2x^2_t}{1+2x_t}$$ The radius of the circle is $$r=\sqrt{(a-x_t)^2+(a-x^2_t)^2} = \left|2 - a\right|$$ A bit of transformation and we get: $$(2x^3_t-2x^2_t)^2+(x_t-x^2_t)^2=(2x^3_t-3x_t-2)^2$$ Unfortunately this equation doesn't have nice solution, so you'll have to calculate numerically Here's what I found on Wolfram Alpha:

$$x_t\approx -0.441229$$; $$x_t\approx 1.33284$$; $$x_t\approx 2.05075$$

Then we can calculate $$a$$ and $$r$$

The circle you showed has $$a \approx 1.6554$$ and $$r \approx 0.34456$$

Let $$Q(b,b^2)$$ and $$P(a,\sqrt a)$$ be the tangency points of the circle on the curves $$y=x^2$$, $$y=\sqrt x$$ respectively. Then the intersection of the corresponding tangent lines is $$R\left(\frac{2\sqrt a\,b^2+a}{4\sqrt a\, b-1},\frac{2ab+b^2}{4\sqrt a\, b-1}\right).$$ Since $$QO=PO$$ where $$O$$ is the center of the circle, we have $$QR=PR$$. Now $$QR^2=PR^2$$ gives $$(2a-2b)(2\sqrt a\,b^2+a)+(2\sqrt a-2b^2)(2ab+b^2)=(a^2+a-b^4-b^2)(4\sqrt a\,b-1).$$ WolframAlpha finds $$b=\sqrt a$$. I am still looking at the equation.

When $$b=\sqrt a$$, we have two solutions of the problem in OP. The smaller circle has the center $$O\approx(1.65544,1.65544)$$ and radius $$r\approx 0.34456$$.