<b>(i) Coordinates of the centre C</b>
Step 1: $AB$ is a diameter, so the centre $C$ is the mid-point of $AB$.
Step 2: Apply the mid-point formula with $A(10,20)$ and $B(50,50)$.
$$C=\left(\frac{10+50}{2},\;\frac{20+50}{2}\right)=\left(\frac{60}{2},\;\frac{70}{2}\right)$$
Step 3: Therefore $C=(30,35)$.

<b>(ii) Radius of the circular park</b>
Step 4: Find $AB$ using the distance formula.
$$AB=\sqrt{(50-10)^{2}+(50-20)^{2}}=\sqrt{40^{2}+30^{2}}$$
Step 5: Simplify.
$$AB=\sqrt{1600+900}=\sqrt{2500}=50\text{ units}$$
Step 6: Radius $=\dfrac{AB}{2}=\dfrac{50}{2}=25$ units.

<b>(iii)(a) Coordinates of P</b>
Step 7: Since $AP=PQ=QB$, the point $P$ divides $AB$ in the ratio $1:2$.
Step 8: Apply the section formula with $m:n=1:2$.
$$P=\left(\frac{1(50)+2(10)}{1+2},\;\frac{1(50)+2(20)}{1+2}\right)$$
Step 9: Simplify.
$$P=\left(\frac{70}{3},\;\frac{90}{3}\right)=\left(\frac{70}{3},\;30\right)$$

<b>OR (iii)(b) Distance of Q from gate A</b>
Step 10: Since $AP=PQ=QB$, the segment $AB$ is divided into 3 equal parts and $Q$ is the second point, so
$$AQ=\frac{2}{3}\times AB$$
Step 11: Substitute $AB=50$ units.
$$AQ=\frac{2}{3}\times 50=\frac{100}{3}=33.33\text{ units}$$
Hence $C=(30,35)$, radius $=25$ units, $P=\left(\dfrac{70}{3},30\right)$ and $AQ=\dfrac{100}{3}$ units.