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

Let the coordinates of D be (p, -2) and F be (x_F, y_F). Since E(6; 2) is the midpoint of DF, we apply the midpoint formula:

$E_x = \frac{x_D + x_F}{2} \, \text{and} \, E_y = \frac{y_D + y_F}{2}$

Substituting E's coordinates gives:

$6 = \frac{p + x_F}{2} \, \text{and} \, 2 = \frac{-2 + y_F}{2}$

From the first equation, solving for $x_F$ gives:

12 = p + x_F \ x_F = 12 - p$$ Substituting $p = 4$ results in: $$x_F = 12 - 4 = 8$$ From the second equation, solving for $y_F$ results in: $$2 = \frac{-2 + y_F}{2} \ 4 = -2 + y_F \ y_F = 6$$ Thus, the coordinates of F are (8, 6).

To find the equation of the tangent line DF, we first calculate the slope (m) between points D(p, -2) and F(8, 6).

Using the slope formula:

$m = \frac{y_F - y_D}{x_F - x_D} = \frac{6 - (-2)}{8 - p} = \frac{8}{8 - p}$

The equation of the line in point-slope form is:

$y - y_1 = m(x - x_1)$

Substituting F's coordinates:

$y - 6 = \frac{8}{8 - p}(x - 8)$

Rearranging to slope-intercept form gives us:

$y = \frac{8}{8 - p}(x - 8) + 6$.

This represents the equation of the common tangent DF.

We can use the distance between the orthogonal tangent and the centre of the larger circle to find the value of $t$.

Using the formula for line distance:

$d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$,

where the coordinates for G are (t, 0) and with calculated values, we derive:

$t = 20$.

Thus, the value of $t$ is 20.

The larger circle's center G is (t; 0) where t = 20. The radius is calculated based on the distance from G to the nearest point on the smaller circle.

Hence, the equation becomes:

$\left(x - 20\right)^2 + y^2 = r^2$,

substituting calculated values results in:

$x^2 + y^2 - 40x + 220 = 0$.

The radius of the smaller circle is $r = \sqrt{5}$ and the larger circle is $R = \sqrt{180}$. The distance between their centers determines $k$:

$d = \text{Radius of larger circle} - \text{Radius of smaller circle} = (6\sqrt{5} - \sqrt{5}) = (5\sqrt{5})$.

Substituting gives the possible values of $k$ as:

$k = \pm 2.5$.