It's a straightforward but messy exercise to find the area of the "lens" formed by the intersection of the two circles. If the radii are the same, consider a circle with center $(0,0)$ and radius 1 and another with center $(d,0)$ and radius $1$. Then the area of the lens divided by $\pi$ is the percentage you are looking for. The circles are
$$C_0: \ \ x^2+y^2 = 1,\ \ \ C_d: \ \ (x-d)^2 +y^2 = 1.$$
Solving for the intersection points we find
$$ P_1 = (d/2,\sqrt{4 - d^2}/2 , \ \ P_2 = (d/2,-\sqrt{4 - d^2}/2) $$
The segment $P_1 P_2$ is a chord that cuts both circles. You can find formulas for the area between each circle and the chord -- then add them to obtain the area of the lens.
The area between the chord and $C_0$ is
$$A_0 = \arccos(d/2)- d\sqrt{4 - d^2}/4.$$
The area between the chord and $C_d$ is identical by symmetry
$$A_d = \arccos(d/2)- d\sqrt{4 - d^2}/4.$$
Hence the percentage of overlap is
$$O = \frac{2 \arccos(d/2)- d\sqrt{1 - d^2/4}}{\pi}.$$
In order to find $d$ for a specified value of $O$, you have to solve a non-linear equation numerically. There is no closed form solution.
If $O = 0.5$ then $d = 0.8079455...$ approximately.
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Let $D,E,F$ be a point on $BC,CA,AB$ such that $AD\perp BC,BE\perp AC,CF\perp AB$ respectively.
Then, we know that $E,F$ exist on the two circles.
Now consider the tangent lines at $F$ for both circles.
Then, let $G$ be a point both on the tangent line for the circle on $AH$ and on $BC$. Also, let $I$ be a point both on the tangent line for the circle on $BC$ and on $AC$.
Now we have
$$\angle{CFG}=\angle{FAH},\quad \angle{IFC}=\angle{FBC}$$
Since
$$\angle{FAH}+\angle{FBC}=\angle{BAD}+\angle{ABD}=90^\circ$$
we have
$$\angle{CFG}+\angle{IFC}=90^\circ.$$
Best Answer
The triangle $ABC$ is isosceles with base $AC$. By construction, the angles $\angle{KCH}$ and $\angle{CBH}$ are congruent, since they are complementary to the two base angles. Therefore, the triangles $\triangle{KCH}$ and $\triangle{CBH}$ are similar, and we have the proportion $KH:CH=CH=BH$.
Now, by simplicity, let us rename the base and the height of the initial triangle as $\overline{AC}=b$ and $\overline{BH}=h$. Also, let us set $\overline{OH}=k$ and $\overline{OK}=x$ (this last segment is what the OP is asking for). Using the proportion above, we get
$$\overline{KH}=\frac{b^2}{4h}$$
and then
$$x=k-\frac{b^2}{4h}$$
The coordinates of the vertices $A,B,C$ can be easily obtained by considering the equations of the three circles and have already been reported in another answer. From this, we have
$$k=\frac{1}{2}\sqrt{\frac{11}{5}}$$ $$b=1-2\sqrt{\frac{11}{5}}$$ $$h=\frac{1}{2}\sqrt{\frac{11}{5}}+\sqrt{3}-1$$
Substituting in the equation above (again by simplicity I omit the calculations and simplifications, but you can check the result here), we get a rather complicated fraction with various radicals that can be written in several ways. Probably the most compact form is
$$x=\frac{\sqrt{55} + \sqrt{165}-19}{ \sqrt{55}+ 10 \sqrt{3} -10}\\ \approx 0.08559789...$$
or alternatively
$$x=\frac{\sqrt{11/5} + \sqrt{33/5}-19/5}{ \sqrt{11/5}+ 2 \sqrt{3} -2}$$
which gives the same numerical value. It is straightforward to express this result in terms of $a$, by taking into account that each square edge in the grid is equal to $a/8$.
Just for a rapid numerical check, approximated to 4 decimal digits, using a different method: the slope of the line $AK$ is given by $h/(b/2)$. Taking $$h= \left(\frac12 \, \sqrt{11/5}+\sqrt{3}−1\right)\approx 1.4737$$ and $$b/2=(1/2−\sqrt{11/5}\approx 0.9832$$ we have that the slope is $$1.4737/0.9832\approx 1.4988$$ The line passes through point $A$, whose $x$-coordinate is $-1/2\sqrt{11/5}\approx-0.7416$ and whose $y$-coordinate is $\approx -0.9832$. The intercept of the line $AK$ is then $$-0.9832+1.4988\cdot 0.7416\approx 0.1283$$
From this, setting $y=0$ in the equation $y=1.4988 x + 0.1283$, we get that the $x$-coordinate of $K$ is $$-0.1283/1.4988\approx -0.0856$$ in good agreement with the exact value calculated above.