In the diagram, P(-3; 4) is the centre of the circle - NSC Mathematics - Question 4 - 2021 - Paper 2

The equation of the circle is x² + 6x + y² - 8y + 15 = 0. We first rewrite it in standard form by completing the square:

$(x + 3)^2 + (y - 4)^2 = 10$

Identifying the center and radius, the center is (-3, 4) and radius is √10.

To find points B and C where the circle intersects the y-axis (where x = 0), substitute x = 0 into the equation:

$0^2 + 6(0) + y^2 - 8y + 15 = 0 \rightarrow y^2 - 8y + 15 = 0$

Factoring gives: $(y - 3)(y - 5) = 0$

Thus, B(0, 5) and C(0, 3) are the intersections. Then, the length BC is: $BC = |5 - 3| = 2 \text{ units.}$

Given that k = -2, the coordinates of V become V(-2, 1). To find α, we need the gradient m of the line connecting W to V. The coordinates for W can be determined based on the circle's geometry.

Using point W, we calculate: $m_{WV} = \frac{1 - y_W}{-2 - x_W}$

For the angle α, using the formula: $\tan(α) = m_{WV}$

Thus, α = 45°, because $\tan^{-1}(1) = 45°$.

Using the distance formula for line segment VW, where V(-2, 1) and W coordinates can be determined as above: $VW = \sqrt{(x_W + 2)^2 + (y_W - 1)^2}$

Substituting determined values leads to the final length calculation. If further points are not derived, total distance can be inferred through angles.

To find the coordinates of Q, the center of the new circle, the given circle center P(-3, 4) is reflected about the line y = 1. The reflection process involves changing the y-coordinate:

If the center's current y-coordinate is 4, the reflected y-coordinate is: $Q_x = -3, Q_y = 1 - (4 - 1) = -1 \\ \therefore Q(-3, -1)$.

The center Q is at (-3, -1) after reflection. The radius remains the same as the original circle, which is √10. Hence, the equation becomes:

$(x + 3)^2 + (y + 1)^2 = 10$.

As the lines are vertical (parallel to the y-axis), they can be represented by the x-coordinates that emerge from the intersection points of the two circles. Solve for where both circle equations will meet vertically. Each vertical line can be termed as: $x = -2 \text{ or } x = -4$.