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 point-slope form of the equation, we have:

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

Substituting m = -\frac{19}{5} and the point S(1, -14):

$y + 14 = -\frac{19}{5}(x - 1)$

Thus, rearranging:

$y = -\frac{19}{5}x + \frac{19}{5} - 14$

$y = -\frac{19}{5}x - \frac{61}{5}$

The center of the circle M has coordinates M(-2, 1), and the radius is the distance from M to P:

$r = \sqrt{(-4 - (-2))^2 + (5 - 1)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$

Therefore, the equation of the circle is:

$(x + 2)^2 + (y - 1)^2 = (2\sqrt{5})^2$

which simplifies to:

$(x + 2)^2 + (y - 1)^2 = 20$

We will use trigonometric ratios in triangle PKR to find the angle. The lengths PK and KR can be calculated:

• For PK:

$PK = \sqrt{(-4 - 0)^2 + (5 - (-3))^2} = \sqrt{16 + 64} = 8$

• For KR:

$KR = \sqrt{(0 - (-4))^2 + ((-3) - 5)^2} = \sqrt{16 + 64} = 8$

Using the cosine rule:

$cos(∠PKR) = \frac{PK^2 + KR^2 - PR^2}{2(PK)(KR)}$

Substituting the values:

$cos(∠PKR) = \frac{8^2 + 8^2 - PR^2}{2(8)(8)}$

Finally, find the angle using the inverse cosine function.

The slope of radius MK is given. Since tangent is perpendicular to the radius:

$m_{tangent} = -\frac{1}{m_{MK}}$

We can find c using the point K(0, -3):

$y + 3 = m_{tangent}(x - 0)$

Thus:

$y = m_{tangent}x - 3$

To find the intersection points, substitute y from the line equation into the circle's equation:

1. Substitute:
   $(x + 2)^2 + \left(\frac{1}{2}x + c - 1\right)^2 = 20$
2. Solve the quadratic equation obtained. This will help determine the conditions for two intersection points.

To calculate the area of triangle SMK, we can use the formula:

$Area = \frac{1}{2} \cdot base \cdot height$

Choosing the base as SM and understanding the height as the distance from K to line SM, we can substitute the appropriate values from previously calculated coordinates to find the area.