In the diagram, the circle with centre O has the equation $x^2 + y^2 = 20$ - NSC Mathematics - Question 4 - 2023 - Paper 2

Let F have coordinates $(x_F; y_F)$. Since E is the midpoint: $E(x_E; y_E) = \left( \frac{p + x_F}{2}; \frac{-2 + y_F}{2} \right)$ From E(6; 2), we have: $\frac{p + x_F}{2} = 6 \quad \text{and} \quad \frac{-2 + y_F}{2} = 2$ Thus:

1. From ( \frac{p + x_F}{2} = 6): $p + x_F = 12 \implies x_F = 12 - p = 12 - 4 = 8$
2. From ( \frac{-2 + y_F}{2} = 2): $-2 + y_F = 4 \implies y_F = 6$ So, F = (8; 6).

The slope of DF can be found using coordinates D(4; -2) and F(8; 6): $m_{DF} = \frac{y_F - y_D}{x_F - x_D} = \frac{6 - (-2)}{8 - 4} = \frac{8}{4} = 2$ Now using point-slope form of the line equation: $y - y_D = m_{DF}(x - x_D)$ Substituting known values gives: $y + 2 = 2(x - 4) \implies y + 2 = 2x - 8 \implies y = 2x - 10$

To find $t$, note that the center G of the larger circle is G(t; 0). We need the distance from G to D and from G to F to be equal, as they are the radii of the circles. Using Pythagorean Theorem: $GA^2 = GD^2 + DF^2$ Where GA = distance from G to A, and D and F coordinates are known. Calculate both distances, set them equal, and solve for $t$. Alternatively, refer to the Pythagorean relationships to directly arrive at: $$ t = 20, $ this depends on substitution and simplification of values found.

With center G(20; 0) and presumed radius (16), the equation takes the form: $(x-20)^2 + y^2 = r^2$ Where radius can be found as seen from previous parts, thus substituting into gives: $(x^2 - 40x + 400) + y^2 - 180 = 0;$ Thus yielding: $x^2 + y^2 - 40x + 220 = 0$.

The smaller circle has radius: $r_{small} = \sqrt{5}$ and larger has radius: $r_{large} = \sqrt{20}$. Touching internally means: $d = r_{large} - r_{small} = k$ Applying values gives: $20 - 4 \implies k = \sqrt{5} - \sqrt{20} = -8 \text{ or } -20;$ thus calculate for k in both cases.