In the diagram below, O(0; 0) is the centre of circle ABC with A(-4; -3) and C(4; -3) - NSC Technical Mathematics - Question 2 - 2019 - Paper 2

Point B lies on the circle. Given the center (0, 0) and radius 5, we determine B's coordinates via the tangent relationship. From the configuration, we see that B is directly above the center along the y-axis:

[ B(0, 5) ]

To find the gradient of line PQ, we first need the coordinates of points P and Q. Once those points are established, the gradient (m) can be calculated using the formula:

[ m = \frac{y_2 - y_1}{x_2 - x_1} ]

Assuming the coordinates of Q are (4, -3), the gradient will depend on the coordinates of P.

With the gradient (m) known, the equation of line PQ can be calculated using the point-slope form:

[ y - y_1 = m(x - x_1) ]

Substituting the coordinates of point P and the gradient will yield the final equation for PQ.

To express the given equation [ x^2 + 8y^2 - 32 = 0 ] in the desired form, we first rearrange it:

[ 8y^2 = 32 - x^2 ] [ y^2 = \frac{32 - x^2}{8} ]

Now, dividing both sides by 32 gives us:

[ \frac{x^2}{32} + \frac{y^2}{4} = 1 ]
(where [ a^2 = 32 \text{ and } b^2 = 4 ])

The resulting equation represents an ellipse centered at the origin. The x-intercepts occur at [ \pm \sqrt{32} ] and the y-intercepts at [ \pm 2 ]. Plot these points to sketch the graph of the ellipse, ensuring that both intercepts are clearly indicated.