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I would continue by replacing $x_P$ and $y_P$ with $2\cos(\theta)$ and $2\sin(\theta)$, respectively. This will give a function $f(\theta)$. Graphing, the minimum is seen at about $82^\circ$, corresponding to a value of approximately $4.086$ for the objective function. If you write the equation in terms of $x=\tan(\theta/2)$ you can take the derivative wrt x and find the minimum. The derivative will have radicals which can be removed and the resulting polynomial does not have a closed form solution (though can easily be solved numerically). Testing the real possibilities to see which minimizes the function confirms the graphical solution.

It is so tempting to think of minimizing $a + b$ as the same as minimizing $a^2 + b^2$ when faced with the square roots of expressions for $a$ and $b$. And the reward, for this problem, is a tractable solution...which is close (at $\theta=-2\tan^{-1}(1/20 - \sqrt{401}/20)\approx 87^\circ$) but is about $5^\circ$ from the correct answer.

If I re-derive the objective function according to the description you gave, I get something else. Here is the derivation using the CAS SymPy:

So the minimum is $\sqrt{41}$ at a value of $t =2\tan^{-1}(8/5 + \sqrt{139}/5)$. This agrees with the graphical minimum of the objective function.

Here is a derivation using SymPy, a CAS in Python:

So the minimum is $\sqrt{41}$ at a value of $\tan(t/2) =(8/5 + \sqrt{139}/5)$. This agrees with the graphical minimum of the objective function.

# define objects

>>> from sympy.abc import *
>>> from sympy import *
>>> P=Circle((5,0),2).arbitrary_point(t)
>>> C=Point(0,5)
>>> A=Point(1,0)

# define objective

>>> objective=Segment(P,C).length + Segment(A,C).length/2; o
sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2) + sqrt(26)/2

# we desire the minimum wrt t (which controls where we are on the circle)

>>> objective.diff(t)
(2*(2*sin(t) - 5)*cos(t) - 2*(2*cos(t) + 5)*sin(t))/sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2)

# rewrite sin and cos in terms of tan and let tan(t/2) be x

>>> _.rewrite(tan).subs(tan(t/2),x).as_numer_denom()[0].factor()
10*(x**2 + 1)*(x**2 - 2*x - 1)

# get solutions for x

>>> solve(_)
[-I, I, 1 - sqrt(2), 1 + sqrt(2)]

# keep real solutions and map back to t

>>> flatten([solve(tan(t/2)-i) for i in _[-2:]])
[-pi/4, 3*pi/4]

# check value of objective at those values of t

>>> [objective.subs(t,i).simplify() for i in _]
[2 + sqrt(26)/2 + 5*sqrt(2), -2 + sqrt(26)/2 + 5*sqrt(2)]
>>> [i.n() for i in _]
[11.6205775686619, 7.62057756866187]

So the minimum is $-2 + \sqrt{26}/2 + 5\sqrt2 \approx 7.6$ and a value of $t =3\pi/4$. This agrees with the graphical minimum of the objective function.

# define objects

>>> from sympy.abc import *
>>> from sympy import *
>>> P=Circle((5,0),2).arbitrary_point(t)
>>> C=Point(0,5)
>>> A=Point(1,0)

# define objective

>>> objective=Segment(P,C).length + Segment(P,A).length/2
sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2) + sqrt((2*cos(t) + 4)**2 + 4*sin(t)**2)/2

# we desire the minimum wrt t (which controls where we are on the circle)

>>> objective.diff(t)
(2*(2*sin(t) - 5)*cos(t) - 2*(2*cos(t) + 5)*sin(t))/sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2) + (-2*(2*cos(t) + 4)*sin(t) + 4*sin(t)*cos(t))/(2*sqrt((2*cos(t) + 4)**2 + 4*sin(t)**2))

# rewrite sin and cos in terms of tan and let tan(t/2) be x, keep only
# the numerator

>>> _.rewrite(tan).subs(tan(t/2),x).as_numer_denom()[0]
(x**2 + 1)**2*(8*x*(1 - x**2) - 8*x*(x**2 + 3))*sqrt((4*x/(x**2 + 1) - 5)**2 + (2*(1 - x**2)/(x**2 + 1) + 5)**2) + (x**2 + 1)**2*(-8*x*(3*x**2 + 7) + 4*(1 - x**2)*(-5*x**2 + 4*x - 5))*sqrt(16*x**2/(x**2 + 1)**2 + (2*(1 - x**2)/(x**2 + 1) + 4)**2)

# remove radicals

>>> unrad(_)
(25*x**22 - 100*x**21 + 339*x**20 - 1440*x**19 + 1991*x**18 - 7020*x**17 + 6949*x**16 - 16320*x**15 + 16346*x**14 - 17640*x**13 + 27454*x**12 + 33614*x**10 + 24360*x**9 + 29866*x**8 + 31680*x**7 + 18749*x**6 + 19980*x**5 + 7871*x**4 + 6560*x**3 + 1979*x**2 + 900*x + 225, [])

# factor

>>> factor(_[0])
(x**2 + 1)**8*(5*x**2 - 16*x - 15)*(5*x**4 - 4*x**3 + 30*x**2 - 44*x - 15)

# get solutions for x

>>> [i for i in solve(_) if i.is_real]
[8/5 - sqrt(139)/5, 8/5 + sqrt(139)/5, 1/5 + sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))/2 + sqrt(-192/25 - 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3) + 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 1616/(125*sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))))/2, -sqrt(-192/25 - 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3) + 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 1616/(125*sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))))/2 + 1/5 + sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))/2]

>>> ans = flatten([solve(tan(t/2)-i) for i in _])

# check value of objective at those values of t

>>> [objective.subs(t,i).n() for i in ans]
[11.3350373276846, 6.40312423743285, 7.11408182438962, 11.9279919892822]

# value of objective at ans[1]

>>> m=objective.subs(t,ans[1]);m
sqrt(4*sin(2*atan(8/5 + sqrt(139)/5))**2 + (2*cos(2*atan(8/5 + sqrt(139)/5)) + 4)**2)/2 + sqrt((2*cos(2*atan(8/5 + sqrt(139)/5)) + 5)**2 + (-5 + 2*sin(2*atan(8/5 + sqrt(139)/5)))**2)

# looks worse that it is - this is equivalent to sqrt(41)

>>> nsimplify(m)
sqrt(41)
>>> m.equals(_)
True

So the minimum is $\sqrt{41}$ at a value of $t =2\tan^{-1}(8/5 + \sqrt{139}/5)$. This agrees with the graphical minimum of the objective function.

If I re-derive the objective function according to the description you gave, I get something else. Here is the derivation using the CAS SymPy:

# define objects

>>> from sympy.abc import *
>>> from sympy import *
>>> P=Circle((5,0),2).arbitrary_point(t)
>>> C=Point(0,5)
>>> A=Point(1,0)

# define objective

>>> objective=Segment(P,C).length + Segment(A,C).length/2; o
sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2) + sqrt(26)/2

# we desire the minimum wrt t (which controls where we are on the circle)

>>> objective.diff(t)
(2*(2*sin(t) - 5)*cos(t) - 2*(2*cos(t) + 5)*sin(t))/sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2)

# rewrite sin and cos in terms of tan and let tan(t/2) be x

>>> _.rewrite(tan).subs(tan(t/2),x).as_numer_denom()[0].factor()
10*(x**2 + 1)*(x**2 - 2*x - 1)

# get solutions for x

>>> solve(_)
[-I, I, 1 - sqrt(2), 1 + sqrt(2)]

# keep real solutions and map back to t

>>> flatten([solve(tan(t/2)-i) for i in _[-2:]])
[-pi/4, 3*pi/4]

# check value of objective at those values of t

>>> [objective.subs(t,i).simplify() for i in _]
[2 + sqrt(26)/2 + 5*sqrt(2), -2 + sqrt(26)/2 + 5*sqrt(2)]
>>> [i.n() for i in _]
[11.6205775686619, 7.62057756866187]

So the minimum is $-2 + \sqrt{26}/2 + 5\sqrt2 \approx 7.6$ and a value of $t =3\pi/4$. This agrees with the graphical minimum of the objective function.