Let y=x be the equation of a chord of the circle C_ 1 (in the...

Circle $C_1$ has radius 5 and chord on line $y = x$ through origin.
Setting center at $(5, 0)$ (in region $x \geq 0$), the chord endpoints are $(0, 0)$ and $(5, 5)$ by solving $(t-5)^2 + t^2 = 25$.
Circle $C_2$ has this chord as diameter, so center is at $(\frac{5}{2}, \frac{5}{2})$ with radius $\frac{5\sqrt{2}}{2}$.
The chord of $C_2$ through $(2, 3)$ that is farthest from center is perpendicular to the radial direction from center to $(2, 3)$.
The radial direction is $(2 - \frac{5}{2}, 3 - \frac{5}{2}) = (-\frac{1}{2}, \frac{1}{2})$ with normal direction $(1, -1)$.
The chord equation is $1(x - 2) - 1(y - 3) = 0$, giving $x - y + 1 = 0$.
In form $x + ay + b = 0$, we have $a = -1$ and $b = 1$, so $a - b = -1 - 1 = -2$.