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 calculated gradient from step 4.1.1, we can use point S (or R) coordinates in the point-slope form of the equation:

$y - y_1 = m(x - x_1)$ Rearranging this into the slope-intercept form (y = mx + c) will yield the equation of SR.

The center of the circle M is the midpoint of P and K. First, calculate the center coordinates:

$M\left(\frac{-4 + 0}{2}, \frac{5 - 3}{2}\right) = M\left(-2, 1\right)$ Next, find the radius (r) which is half the distance PK:

$r = \frac{1}{2} \sqrt{(-4 - 0)^2 + (5 + 3)^2} = \frac{1}{2} \sqrt{16 + 64} = \frac{1}{2}\cdot \sqrt{80} = 4\sqrt{5}$ Thus, the equation of the circle is:

$(x + 2)^2 + (y - 1)^2 = 80$

Using point K and the previously determined angles, we can find the lengths PK and KR. Using trigonometric functions:

• Calculate the lengths PK and KR.
• Use the cosine rule:

$\cos PKR = \frac{PK^2 + KR^2 - PR^2}{2 \cdot PK \cdot KR}$ where PKR is the angle formed. The final angle can be calculated using:

$\theta = \tan^{-1}\left(\frac{y}{x}\right)$

To find the tangent at point K, use the gradient of the radius MK. The gradient of the tangent (m) is the negative reciprocal of the radius:

$m = -\frac{1}{\text{gradient of MK}}$ Using point-slope form with point K's coordinates will give us the tangent equation.

Substitute the line equation into the circle's equation:

Substituting for y yields a quadratic in x. To ensure two different points of intersection, the discriminant should be positive:

$D > 0$ Analyze the resulting quadratic equation to determine the allowable values of t.

To find the area of the triangle ΔSMK, use:

$Area = \frac{1}{2} \cdot base \cdot height$ Determine the coordinates of S, M, and K and calculate the lengths SK and SM to find the area using the determinant method or base-height method.