Can you solve using a single equation?

The centre must lie on the perpendicular bisectors of the two chords in order for it to be equidistant from the four given intercepts (from the basic definition of a circle, assuming this is indeed a circle).

The perpendicular bisectors are the lines $x = 6$ and $y = 6$ , hence the centre is $(6, 6)$ .

Thus we can write the equation of the circle thus: $(x - 6)^2 + (y - 6)^2 = R^2.$

Substituting one of the intercepts: $(14 - 6)^2 + (0 - 6)^2 = R^2 \\ \Rightarrow 8^2 + 6^2 = R^2 \\ \Rightarrow 100 = R^2 \\ \Rightarrow R = \boxed{10}.$

If you look to see how the circle would have to shift to be centered on the origin we can see that it will need to shift 6 to the left and 6 down. Using this information we can show that the right triangle created using the radius and the shift of 6 and the remaining height of 8 creates a 3-4-5 triangle making the radius 10.

Label the points as follows:

If two chords intersect in the interior of a circle, then the measure of each angle is one half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Therefore, $\angle AOB = \frac{1}{2}(m \widehat{AB} + m \widehat{CD}) = 90°$ . By symmetry, $m \widehat{AB} = m \widehat{CD}$ , so that this solves to $m \widehat{AB} = 90°$ . Since the measure of a minor arc is also defined as the measure of its central angle, $\angle APB = m \widehat{AB} = 90°$ .

Therefore, $\triangle AOB$ and $\triangle APB$ are right triangles sharing the same hypotenuse, so that by Pythagorean's Theorem, $AB^2 = AO^2 + BO^2 = AP^2 + BP^2$ , or

$14^2 + 2^2 = 2R^2$

and this single equation solves to $R = \boxed{10}$ .

Let the green and purple chords coincide with the x & y-axes respectively (such that their intersection point is the origin). Let the center of this circle be at the point $(\alpha,\beta)$ such that the required equation becomes $(x-\alpha)^2 + (y-\beta)^2 = R^2.$ Substituting the four given circumferential points: $(x,y) = (14,0), (-2,0), (0,14), (0, -2)$ above yields the following system of equations:

$(14-\alpha)^2 + (0-\beta)^2 = R^2;$

$(-2-\alpha)^2 + (0-\beta)^2 = R^2;$

$(0-\alpha)^2 + (14-\beta)^2 = R^2;$

$(0-\alpha)^2 + (-2-\beta)^2 = R^2$

which ultimately produce the center point $(\alpha,\beta) = (6,6)$ and $(x-6)^2 + (y-6)^2 = R^2$ . Picking any one of the four given circumferential points finally produces $\boxed{R = 10}.$

Draw two lines

connecting A : (-2,0) between B : (0,14) and B : (0,14) between C : (14,0)

and then use the law of sine:

AB/ sin(<BCA) = 2R

length of AB : root(14^2 + 2^2) = 10*root(2)

<BCA : pi/4

sin(<BCA) = 1/(root(2))

---

10*root(2)/{(1/(root(2))} =2R

=>R=10