Let y=x be the equation of a chord of the cir | vedclass

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(C) The chord $y=x$ is a diameter of $C_{2}$. The endpoints of this chord on $C_{1}$ are $(0,0)$ and $(5,5)$ because the distance between them is $\sq

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-$2$

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(C) The chord $y=x$ is a diameter of $C_{2}$. The endpoints of this chord on $C_{1}$ are $(0,0)$ and $(5,5)$ because the distance between them is $\sqrt{5^2+5^2} = \sqrt{50} = 5\sqrt{2}$, which is not the diameter. Wait, the chord $y=x$ passes through the origin and has length $L$. Since it is a chord of $C_1$ (diameter $10$), its endpoints are $(0,0)$ and $(5,5)$ is incorrect. The chord $y=x$ passes through the origin. Let the endpoints be $(x_1, x_1)$ and $(x_2, x_2)$. The distance is $\sqrt{2}|x_1-x_2|$. The circle $C_1$ has diameter $10$ and passes through $(0,0)$. The chord $y=x$ is a diameter of $C_2$. The center of $C_2$ is the midpoint of the chord. The chord of $C_2$ passing through $(2,3)$ and farthest from the center $A(\frac{5}{2}, \frac{5}{2})$ is perpendicular to the line segment joining the center $A$ and the point $(2,3)$.Slope of $AB = \frac{3 - 5/2}{2 - 5/2} = \frac{1/2}{-1/2} = -1$.Since the required chord is perpendicular to $AB$, its slope is $m = -1/(-1) = 1$.The equation of the chord is $y - 3 = 1(x - 2)$, which simplifies to $x - y + 1 = 0$.Comparing this with $x + ay + b = 0$, we get $a = -1$ and $b = 1$.Therefore, $a - b = -1 - 1 = -2$.

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