In the diagram, the circle centred at C(2; p) is drawn - NSC Mathematics - Question 4 - 2024 - Paper 2

The distance between points A(5, 1) and B(-3, -3) can be calculated using the distance formula:

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

Substituting in the coordinates:

$AB = \sqrt{(-3 - 5)^2 + (-3 - 1)^2} = \sqrt{(-8)^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$

First, calculate the gradient (slope) of AB:

$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 1}{-3 - 5} = \frac{-4}{-8} = \frac{1}{2}$

The gradient of the perpendicular bisector is the negative reciprocal:

$m_{CE} = -2$

Using the coordinates of E(1, -1) and the point-slope form:

$y - y_1 = m(x - x_1) \Rightarrow y + 1 = -2(x - 1) \Rightarrow y = -2x + 1$

Substituting into the circle's equation, where the center C is (2, p):

Since E(1, -1) is on the perpendicular bisector:

Line equation simplifies to:

$y = -2x + 1$

Substituting x = 2 into the equation:

$-1 = -2(1) + c \Rightarrow c = 1 \Rightarrow p = -3$

The general form of the circle's equation is:

$(x - h)^2 + (y - k)^2 = r^2$

Where C(2, p) and r = 5 units:

Expanding the equation:

$(x - 2)^2 + (y - p)^2 = 25\n\Rightarrow x^2 - 4x + 4 + y^2 - 2py + p^2 = 25 \n\Rightarrow x^2 + y^2 - 4x - 2py + (p^2 - 21) = 0$

By substituting p = -3, we get:

$x^2 + y^2 - 4x + 6y - 12 = 0$. This confirms the given equation.

Substituting y = tx + 8 into the circle's equation:

$x^2 + (tx + 8)^2 - 4x + 6(tx + 8) - 12 = 0 \Rightarrow A, B, C = 0$

For no intersection, the discriminant must be negative:

$\Delta < 0 \Rightarrow 24 < 3t < 24 \; \Rightarrow \frac{24}{3} > t > \frac{24}{3} \Rightarrow t < 8 \text{ or } t > 8$