Let \(y=x\) be the equation of a chord of the circle \({C}_{1}\) (in the closed half-plane\(x\geq 0\)) of diameter \(10\…

The circle \(C_1\) has a diameter of \(10 \), so its radius is \(R=5\). It passes through the origin \((0,0)\) and is contained in the closed half-plane \(x \geq 0\).

Therefore centre is \((5,0)\)

The equation of \(C_1\) is:

\((x-5)^2+y^2=25\)

\(\Rightarrow x^2+y^2-10 x=0\)

The line \(y=x\) is a chord of \(C_1\). We find the intersection points of \(y=x\) and \(C_1\) :

\((x-5)^2+x^2=25 \)

\( \Rightarrow x^2-10 x+25+x^2=25 \)

\( \Rightarrow 2 x^2-10 x=0 \)

\( \Rightarrow 2 x(x-5)=0\)

The intersection points are \((0,0)\) and \((5,5)\). These are the endpoints of the diameter of circle \(C_2\).

The center \(M\) of \(C_2\) is the midpoint of the chord:

\(M\left(\frac{5}{2}, \frac{5}{2}\right)\)

Given point is \(P(2,3)\)

In any circle, the chord passing through a point \(P\) that is farthest from the center is the one perpendicular to the radius (or segment) \(M P\).

Now the slope of the segment \(M P\) is:

\(m_{M P}=\frac{3-2.5}{2-2.5}=\frac{0.5}{-0.5}=-1\)

The slope \(m\) of the required chord is

\(m=-\frac{1}{-1}=1\)

Now equation of required chord is \(y-3=1(x-2)\)

\(\Rightarrow y-3=x-2 \Rightarrow x-y+1=0\)

we get \( a=-1\) and \( b=1\)

Now \(a-b=-1-1=-2\)