A complicated circle inside a square

After drawing a diagram, I went straight onto an algebraic approach, which I doubt is the best of approaches as it led to a nasty quadratic. $$ I would greatly appreciate it if someone here could suggest another solution. $$ Thank you.

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With point $A$ in the upper-left corner and $ABCD$ clockwise, I first let $D$ have coordinates $(0,0)$ , and the positive $x$ -axis be in the direction of $DC$ . The line $DM$ therefore has equation $5y=12x$ and the circle $(x-6)^{2}+(y-6)^{2}=36$ , which we shall call equations $(1)$ and $(2)$ respectively.

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Replacing $y$ with $\frac{12}{5}x$ in $(2)$ leads us, after simplification, to the nasty quadratic I was on about in the beginning:

$169x^{2}-1020x+900=0$ ,

leading us to the solutions

$x=\frac{510\pm60\sqrt{30}}{169}$ .

Looking back at our diagram, the ' $+$ ' solution corresponds to $P$ , while the ' $-$ ' solution corresponds to $Q$ . We're after $P$ , so take the ' $+$ ' solution.

$y$ is simply $\frac{12}{5}x$ , so

$y=\frac{1224+144\sqrt{30}}{169}$ .

$M$ has coordinates $(5,12)$ , so the distance $MP$ is:

$MP=\sqrt{(5-x)^{2}+(12-y)^{2}}$ $\;\;\;\;\;$ (See 'Details:' below (written after final boxed answer))

$MP=\frac{67-12\sqrt{30}}{13}$

Therefore, the final answer appears:

$a+b+c+d=67+12+30+13=\boxed{122}$

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Details:

After some tedious algebra (I didn't want to use WolframAlpha as I wanted to test myself),

$MP=\frac{1}{13}\sqrt{8809-1608\sqrt{30}}$

This is not in the form required, so I was convinced this was a perfect square.

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Consider $(a-b\sqrt{30})^{2}$ :

$=a^{2}-2ab\sqrt{30}+30b^{2}$

Comparing it with $8809-1608\sqrt{30}$ ,

$a^{2}+30b^{2}=8809$ , (Equation 3)

$ab=804=2^{2}\times3\times67$ (Equation 4)

Since the square of any real number $\geq0$ ,

$a^{2}\leq8809$ and $30b^{2}\leq8809$

Looking at the factorised form of Equation 4, the only integers $a$ and $b$ which satisfy both equation 4 and the inequalities are $a=67$ and $b=12$ .

After confirming that $(67-12\sqrt{30})^{2}$ does equal $8809-1608\sqrt{30}$ , the answer surfaces:

$MP=\frac{67-12\sqrt{30}}{13}$ , and as before,

$a+b+c+d=67+12+30+13=\boxed{122}$