In the diagram below, the equation of the circle with centre O is $x^2 + y^2 = 20$ - NSC Mathematics - Question 4 - 2016 - Paper 2

To find the coordinates of R, we substitute the x-coordinate of R (which is 2) into the circle's equation:

$x^2 + y^2 = 20$

Substituting $x = 2$:

$2^2 + y^2 = 20 \Rightarrow 4 + y^2 = 20 \Rightarrow y^2 = 16 \Rightarrow y = \pm 4$

Thus, the coordinates of R are (2, -4).

To find the area of triangle OTS, we need the lengths OS and OT:

Using point R(2, -4), we find OT by substituting R into the tangent equation:

$-4 = \frac{1}{2}(2) + k \Rightarrow -4 = 1 + k \Rightarrow k = -5$

Now substituting k into the equation of the tangent:

$y = \frac{1}{2}x - 5$

To find where it intercepts the x-axis (S), we set y = 0:

$0 = \frac{1}{2}x - 5 \Rightarrow \frac{1}{2}x = 5 \Rightarrow x = 10 \Rightarrow S(10, 0)$

To find OT, we set x = 0 in the tangent equation:

$y = \frac{1}{2}(0) - 5 \Rightarrow y = -5 \Rightarrow T(0, -5)$

The area of triangle OTS can be calculated using:

$\text{Area} = \frac{1}{2} \times \text{OS} \times \text{OT}$

Where OS = distance from O(0,0) to S(10,0) = 10 units and OT = distance from O(0,0) to T(0,-5) = 5 units:

$\text{Area} = \frac{1}{2} \times 10 \times 5 = 25 \text{ square units}.$

To find the length of VT, we first find the coordinates of point V. Given V(2, -4) and T(0, -5), we will use the distance formula:

$VT = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting the coordinates:

$VT = \sqrt{(0 - 2)^2 + (-5 - (-4))^2} = \sqrt{(-2)^2 + (-5 + 4)^2} = \sqrt{4 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$

Thus, the length of VT is:

$VT = \sqrt{5}.$