Two Circles?

Without any algebra, notice that the first is a circle centered at $(5,4)$ and radius $3$ . Then we want to find the largest radius of a circle centered at $(4,5)$ that touches the other circle. With this geometric interpretation, it's easy to see that this occurs on the line connecting the two centers of the circles. In terms of coordinates, this occurs where $y=9-x$ , which quickly gives the point $\left(\frac{10+3\sqrt2}2,\frac{8-3\sqrt2}2\right)$ .

Solution 1 -We have

$(x-4)^2+(y-5)^2=(x-5)^2+(y-4)^2+2x-2y=9+2(x-y)$ .... So we actually want to maximize $x-y$ .

We'll use substitution,

$(y-4)^2=3^2-(x-5)^2 = (3+x-5)(3-x+5)=-(x-2)(x-8)\\ \therefore y-4=\sqrt{-x^2+10x-16} \implies y =\sqrt{-x^2+10x-16}+4$

Maximize $x-\sqrt{-x^2+10x-16}-4$ , thus $\dfrac{d(x-\sqrt{-x^2+10x-16}-4)}{dx}=0 \implies 1-\dfrac{-2x+10}{2\sqrt{-x^2+10x-16}}=0 \\ \implies \dfrac{x-5}{\sqrt{-x^2+10x-16}}=-1 \\ \therefore (x-5)^2=-x^2+10x-25 \implies 2x^2-20x+41=0$

(further part of this is in solution 2)

---

Solution 2- the wanted expression is

$(x-4)^4+(y-5)^2 \\ = x^2-8x+16 + (\sqrt{-x^2+10x-16}-1)^2 \\ =x^2-8x+16 -x^2+10x-16 +1 -2\sqrt{-x^2+10x-16} \\ = 2\sqrt{-x^2+10x-16} +2x+1$

---

This is maximum at $\dfrac{d(2\sqrt{-x^2+10x-16} +2x+1)}{dx} =0$

---

Thus $2\dfrac{-2x+10}{2\sqrt{-x^2+10x-16}}+2=0 \\ \therefore 2\sqrt{-x^2+10x-16}= -2x+10 \\ \therefore -x^2+10x-16=(x-5)^2=x^2-10x+25 \implies 2x^2-20x+41=0 \\ \therefore 2x^2-20x+32=-9 \implies x^2-10x+16=\dfrac{-9}{2}$

---

And, larger solution of $x$ from this quadratic is $\dfrac{20+\sqrt{400-4\times 2 \times 41}}{4} = 5 + \dfrac{3\sqrt{2}}{2}$

---

Put these values in required equation to get

$max(x-4)^2+(y-5)^2 = 2\sqrt{\dfrac{9}{2} } + 10 + 3 \sqrt{2} +1 = 11 + 6\sqrt{2} \approx \boxed{19.485}$

The distance between (5, 4) and (4, 5) is $\sqrt{2}$ . So, the distance between the points on the circle (determined by the equation) and (4, 5) can have values between $3+\sqrt{2}$ and $3-\sqrt{2}$ . Our goal is to maximize the distance squared which is $(3+\sqrt{2})^2=11+6\sqrt{2}$ .

This question is the same as "if (x,y) is 3 units away from (5,4), what is its maximum distance from (4,5)".

Draw it on a Cartesian plane and you'll get $3+\sqrt2$ units, and the answer is it squared.

With the substitution $x = 3 \sin{x} + 5$ and $y = 3 \cos{x} + 4$ , we see that the initial equation is satisfied. The one we want to maximize can be rewritten as $(3 \sin{x} + 1)^2 + (3 \cos{x} - 1)^2 = 9(\sin^2 {x} + \cos^2 {x}) + 6(\sin{x} - \cos{x}) + 2$ $= 11 + 6 (\sin{x} + \cos{x})$ .

Factoring $\sin{x} + \cos{x}$ to $\sqrt{2} ( \frac{\sqrt{2}}{2} \cdot \sin{x} + \frac{\sqrt{2}}{2} \cdot \cos{x})$ , we see that it equals $\sqrt{2} \cdot \sin{(x+\frac{\pi}{4})}$ , and its maximum is obviously $\sqrt{2}$ .

So our desired value of $11 + 6 (\sin{x} + \cos{x} )= (x-4)^2 + (y-5)^2$ is $11 + 6 \sqrt{2} \approx \boxed{19.485.}$