According to question 
PR = RQ or R is mid-point of PQ.
Using mid-point formula 
\(x=\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 }\ and\ y=\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \)
\(\frac { b-2+4 }{ 2 } =-3\Rightarrow b+2=-6\Rightarrow b=-8\)

A(-6,7),B(-1,-5)
Let x₁=-6,y₁=7;x₂=-1,y₂=-5
Distance, \(AB=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { { 5 }^{ 2 }+\left( -12 \right) ^{ 2 } } =\sqrt { 25+144 } =\sqrt { 169 } =13\)
2AB=2x13=26

True because distance between centre (origin) and Q(6,8) is greater than its radius. i.e., 5

Since the point P(x,y) is equidistant from the points A(7, 1) and B(3, 5). 
Therefore PA = PB ...(i)
Using distance formula
\(PA=\sqrt { \left( x-7 \right) ^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(=\sqrt { { x }^{ 2 }+{ 7 }^{ 2 }-2.7x+{ y }^{ 2 }+{ 1 }^{ 2 }-2.y.1 } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }+49+1-14x-2y } \)
\(PB=\sqrt { \left( x-3 \right) ^{ 2 }+\left( y-5 \right) ^{ 2 } } \)  (using distance formula)
\(=\sqrt { { x }^{ 2 }+{ 3 }^{ 2 }-2.3x+{ y }^{ 2 }+{ 5 }^{ 2 }-2.5.y } =\sqrt { { x }^{ 2 }+9-6x+{ y }^{ 2 }+25-10y } \)
Substituting the values of PA and PB in (i). we get
\(\sqrt { { x }^{ 2 }+{ y }^{ 2 }+50-14x-2y } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }-6x-10y+34 } \)
Squaring both sides, We get
\({ x }^{ 2 }+{ y }^{ 2 }+50-14x-2y={ x }^{ 2 }+{ y }^{ 2 }-6x-10y+34\)
⇒ 50-34=14x+2y-6x-10y ⇒16=8x-8y ⇒x-y=2