In the diagram, the equation of the circle centred at N(-12; 5) is $x^2 + y^2 + 24x - 10y + 153 = 0$ - NSC Mathematics - Question 4 - 2023 - Paper 2

The radius of the smaller circle can be calculated using the equation of the circle at M. The equation is given by $(x + 6)^2 + (y + 3)^2 = 25$.

Here, $r^2 = 25$, therefore the radius $r = 5$ units.

To find the length of TS, we first find the distance between the points S and T. Since T is on the line connecting N and M, find the coordinates of S using the distance formula between N and M (10 units) and the radius (5 units). Thus, the length TS = 1 unit.

Since PR is parallel to the x-axis, the equation can be given by $y = -8$.

To find the slope (m) of PS, we determine the slope of the line connecting M(6, -3) and K(17, -5):

$m_{MK} = \frac{-5 - (-3)}{17 - 6} = -\frac{2}{11}$.

Using point-slope form, the equation is derived as $y + 3 = -\frac{2}{11}(x + 6)$. Simplifying gives $y = -\frac{2}{11}x - \frac{31}{11}$.

To calculate the perimeter, first find the lengths of PR, PS, and MR:

$PR = 5 + 5 + 5 = 15$ units, and $MR = 5$ units. Thus, perimeter = $15 + 5 + 5 = 40$ units.

Using the formula for area with an approximated base and height:

$Area \ of \ ANPS = \frac{1}{2} N.S.SP = \frac{1}{2}(5)(5) = \frac{25}{2}$.

The area of quadrilateral PSMR can be calculated by arrangement or congruency methods. Ultimately, the ratio rac{area \ of \ ANPS}{area \ of \ quadrilateral \ PSMR} is calculated.