In the diagram, the equation of the circle with centre F is $(x - 3)^2 + (y - 1)^2 = r^2$ - NSC Mathematics - Question 4 - 2018 - Paper 2

To find the length of FS, we use the distance formula: $FS = \sqrt{(6 - 3)^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$

Since FH:HG = 1:2, we have HG = 2 imes FH. Given that FH is the radius of the smaller circle, we established that HG = 10.

JH and JK are tangents from the point J to the circles with centres F and G, respectively, thus JH = JK by the tangent-secant theorem.

We know:

1. JK = 20.
2. Since JK is a tangent to the circle at K and JH is also a tangent from J to the circle at H, we have FS which is perpendicular to JK. Thus, using the Pythagorean theorem: $FJ^2 = FS^2 + JK^2\Rightarrow FJ^2 = 5^2 + 20^2 = 25 + 400 = 425\Rightarrow FJ = \sqrt{425} = 5\sqrt{17}.$

The equation of the circle with centre G is given by: $(x - m)^2 + (y - n)^2 = r^2.$ To find parameters m and n, we utilize the distances previously calculated.

Given that JK is tangent to the circle G, we know that at K(22, n), since GK = HG = 10, we find:

1. Therefore, [(22 - m)^2 + (n - n)^2 = 10^2\Rightarrow (22 - m)^2 = 100, m = 22 \pm 10, m = 32 \text{ or } 12.]
2. Now, substituting the valid range for (m): [n = 12,\text{ or } n = 13 \Rightarrow G(22, n) = (22, 12) \text{ or } (22, 13).]