In the diagram, P(-4 ; 5) and K(0 ; -3) are the end points of the diameter of a circle with centre M - NSC Mathematics - Question 4 - 2017 - Paper 2

Using the gradient calculated in 4.1.1 and the point S(1, 7):

The equation can be derived using
$y - y_1 = m(x - x_1)$

Substituting the values:
$y - 7 = \frac{7}{15}(x - 1)$

Expanding gives:
$y = \frac{7}{15}x + \frac{7}{15} + 7$

Therefore, the equation of line SR is (y = \frac{7}{15}x + \frac{122}{15}).

Given that the center M of the circle is at the midpoint of P and K:

M can be calculated as
$M = \left(\frac{-4 + 0}{2}, \frac{5 - 3}{2}\right) = (\-2, 1)$

The radius r can be determined as half the distance between P and K:

$r = \frac{1}{2} \sqrt{(-4 - 0)² + (5 + 3)²} = \frac{1}{2} \sqrt{16 + 64} = \frac{1}{2}\sqrt{80} = 4\sqrt{5}$

Thus, the equation of the circle is:
$(x + 2)² + (y - 1)² = (4\sqrt{5})² = 80$.

To calculate the angle ∠PKR, we can use the tangent ratio:

$\tan(θ) = \frac{opposite}{adjacent}$
Where the lengths of PK and KR are calculated as follows:
$PK = \sqrt{(-4 - 0)^2 + (5 + 3)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$
Also, from angle calculations in triangle,
$\tan(θ) = \frac{PK}{KR}$
Therefore,
$θ = \tan^{-1} \left(\frac{PK}{KR}\right)$.

The gradient of the line tangent at K is the negative reciprocal of the radius slope. If the gradient of the radius from O to K (0, -3) is determined first, the tangent will have the gradient:

$m_t = -\frac{1}{\text{slope of radius}}$
Using this with the point (0, -3),
$y - (-3) = m_t(x - 0)$
Which simplifies to provide the tangent's equation.

To determine the values of t, substitute y = \frac{1}{2}x + t into the circle's equation, and solve for discriminant conditions.

The circle's equation is given by
$(x + 2)^2 + (\frac{1}{2}x + t - 1)^2 = 80$
The equation must yield a positive discriminant to ensure two intersections.

The area of triangle ΔSMK can be found using the formula:

$\text{Area} = \frac{1}{2} \times base \times height$
Identify points S, M, K to find base and height, or use coordinate geometry methods such as:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2|$
where (x1,y1), (x2,y2), (x3,y3) are coordinates of S, M, K respectively.