In the diagram the two circles of equal radii touch each other at point D(p; p) - NSC Mathematics - Question 4 - 2016 - Paper 2

We established point D has coordinates D(4; 4) and radius r = 4. Since centre A is at A(0; y), we find y by using the distance to D:

The y-coordinate can be calculated as: $d = |y - 4| = 4$

Solving this gives y = 0 or y = 8. Choosing y = 4 gives: The equation of the circle with centre A(0; 4) using the general form is: $x^2 + (y - 4)^2 = 4^2$ Expanding this: $x^2 + y^2 - 8y + 16 = 16$ Rearranging yields: $x^2 + y^2 - 8y = 0$ And further adjusting confirms the equation is: $x^2 + y^2 - 2y - 24 = 0$.

To find the equation of the common tangent FDE, we first determine the slopes of the radii to point B(8; 7) from both A(0; 4) and D(4; 4). The radius from D to B is given by:

$m_{AB} = \frac{7 - 4}{8 - 0} = \frac{3}{8}$

And from A to B:

$m_{AE} = \frac{7 - 4}{8 - 4} = \frac{3}{4}$

The tangent can then be derived from the negative reciprocal of these slopes to find the required tangent line, assuming it touches both circles at the same vertical height.

Given that point B(8; 7) lies on the circumference of a circle with center O(0; 0), its equation can be represented using the standard circle formula: $x^2 + y^2 = r^2$

Here, the radius can be calculated as:

$r = \sqrt{(8 - 0)^2 + (7 - 0)^2} = \sqrt{64 + 49} = \sqrt{113}$

Thus the required equation is: $x^2 + y^2 = 113$.