Circles Part 2

Concept:The chord farthest from the center of a circle through a given point is perpendicular to the line joining the center to that point.Explanation:First, find circle $C_1$.Its diameter is 10, so radius $R = 5$.It passes through origin $(0,0)$ and lies in $x \ge 0$.Hence its center is at $(5,0)$ because distance from center to $y$-axis equals radius.Equation of $C_1$: $(x-5)^2 + y^2 = 25$.Chord $y = x$ of $C_1$ intersects at $(0,0)$ and $(5,5)$.These are the endpoints of the diameter of $C_2$.Center $M$ of $C_2$ is midpoint: $(2.5, 2.5)$.We need the chord of $C_2$ through $P(2,3)$ that is farthest from $M$.This chord is perpendicular to $MP$.Slope of $MP$: $\frac{3-2.5}{2-2.5} = \frac{0.5}{-0.5} = -1$.So slope of chord is the negative reciprocal: $1$.Using point-slope through $P$: $y - 3 = 1(x - 2)$.Simplify: $x - y + 1 = 0$.Given chord equation is $x + a y + b = 0$.Comparing: $a = -1$, $b = 1$.Then $a - b = -1 - 1 = -2$.Answer:$-2$ (Option D).