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

Substituting the coordinates of point R into the equation of the circle:

1. The circle equation is:

   $x^2 + y^2 = 20$

2. Let the coordinates of R be (x, y), substituting gives:

   $x^2 + (-4)^2 = 20$

   Therefore,

   $x^2 + 16 = 20$

   This simplifies to:

   $x^2 = 4$

   Thus:

   $x = 2$

3. Therefore, the coordinates of R are:

   $R(2, -4)$

To find the area of triangle OTS:

1. To find OT and OS, we first need to find the slope of line PRS:

   Substitute R into PRS equation:

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

   This results in:

   $k = -5$

   Therefore, the equation of PRS is:

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

2. Finding points T and S:

   • For T (y-axis, x=0):

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

     Thus, T(0, -5)

   • For S (x-axis, y=0):

     $0 = \frac{1}{2}x - 5$

     Solving gives:

     $x = 10$

     Thus, S(10, 0)

3. The area of triangle OTS can be calculated:

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

   Where:

   • $OS = 10$ and $OT = 5$

   Therefore:

   $ext{Area} = \frac{1}{2} \times 10 \times 5 = 25 ext{ sq units}$

To find the length of VT:

1. Calculate the coordinates of T and V:

   • We have T(0, -5) from previous calculation.
   • V corresponds to the y-intercept of OR at (0, 0).

2. Use the distance formula between points V(0, 2) and T(0, -5):

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

   Simplifying gives:

   $VT = \sqrt{(2 + 5)^2} = \sqrt{7^2} = 7$

Thus, the length of VT is:

$VT = 7$