The diagram shows three points A(-16, 0), B(0, 12) and C(8, 2). A circle is drawn with centre C such that AB is a tangent to the circle. The perpendicular distance of a point $(p, q)$ from a line $ax + by + c = 0$ is given by the formula Distance = $\dfrac{|ap + bq + c|}{\sqrt{a^{2} + b^{2}}}$.

1. Find the gradient of line AB from the coordinates of A and B.

For $A(-16,0)$ and $B(0,12)$,

$m=\frac{12-0}{0-(-16)}=\frac{12}{16}=\frac{3}{4}.$

2. Use the gradient and a point on the line to form the equation, then rearrange to the required form.

Using point $B(0,12)$,

$y=\frac{3}{4}x+12.$

Rearranging,

$4y=3x+48$

$3x-4y+48=0.$

1. Substitute the centre C into the perpendicular distance formula using the equation of AB.

From part (i), line $AB$ is
$3x-4y+48=0.$
Here, $a=3$, $b=-4$, $c=48$ and $C=(8,2)$.

Radius
$=\frac{|3(8)+(-4)(2)+48|}{\sqrt{3^2+(-4)^2}}$

2. Evaluate the distance to obtain the radius.

$\frac{|24-8+48|}{\sqrt{9+16}}=\frac{64}{5}=12.8$

Since $AB$ is tangent to the circle, this perpendicular distance from $C$ to $AB$ is the radius.