Let a circle C have its centre in the first quadrant, intersect the...

Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.

Since the circle intersects the coordinate axes at exactly three points, it must pass through the origin and intersect each axis at one other distinct point. Thus, $c = 0$.

The lengths of the intercepts on the x-axis and y-axis are $2\sqrt{g^2 - c} = 2|g|$ and $2\sqrt{f^2 - c} = 2|f|$.

Given that the intercepts are equal, we have $|g| = |f|$.

The centre of the circle is $(-g, -f)$. Since it lies in the first quadrant, $-g > 0$ and $-f > 0$. Therefore, $g = f = -a$ for some $a > 0$.

The centre is $(a, a)$ and the radius is $r = \sqrt{a^2 + a^2} = a\sqrt{2}$.

The distance $d$ from the centre $(a, a)$ to the line $x + y - 1 = 0$ is given by:

$d = \dfrac{|a + a - 1|}{\sqrt{1^2 + 1^2}} = \dfrac{|2a - 1|}{\sqrt{2}}$

The length of the chord on the given line is $2\sqrt{r^2 - d^2} = \sqrt{14}$.

Squaring both sides, we get:

$4(r^2 - d^2) = 14 \Rightarrow r^2 - d^2 = \dfrac{7}{2}$

Substituting $r^2 = 2a^2$ and $d^2 = \dfrac{(2a - 1)^2}{2}$:

$2a^2 - \dfrac{(2a - 1)^2}{2} = \dfrac{7}{2}$

$4a^2 - (4a^2 - 4a + 1) = 7$

$4a - 1 = 7 \Rightarrow 4a = 8 \Rightarrow a = 2$

The square of the radius of the circle is $r^2 = 2a^2 = 2(2)^2 = 8$.

Answer: $8$