In the diagram, the circle centred at M(a ; b) is drawn - NSC Mathematics - Question 4 - 2022 - Paper 2

Given that M lies on the line y = x + 1, we can substitute the coordinates:

If M has coordinates (a, b), then:

$b = a + 1$

We can substitute this in to find M if a is known. In this case, if we take a = 2:

$b = 2 + 1 = 3$

Thus, the coordinates of M are (2, 3).

To calculate the radius, we can use the distance formula between the center M(2, 3) and the tangent point K(5, 7):

$r = ext{distance}(M, K) = \sqrt{(5 - 2)^2 + (7 - 3)^2}$

Calculating this gives:

$r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

To find TR, we calculate the distance between T(0, y_T) and R(6, 0).

Given that M(2; 3) is the center, the y-coordinate of T is:

$y_T = -\frac{3}{5}(0 - 2) + 3 = \frac{6}{5} + 3 = \frac{21}{5}$

Thus, the length of TR can be calculated:

$TR = \sqrt{(6 - 0)^2 + (0 - \frac{21}{5})^2}$

Calculating this will yield:

$TR = 8 \text{ units}$

The slope of the line through M(2, 3) to K(5, 7) is:

$m_{MK} = \frac{7 - 3}{5 - 2} = \frac{4}{3}$

The slope of the tangent (perpendicular to MK) is:

$m_{tangent} = -\frac{3}{4}$

Using point-slope form at K:

$y - 7 = -\frac{3}{4}(x - 5)$

Rearranging gives:

$y = -\frac{3}{4}x + \frac{15}{4} + 7 \Rightarrow y = -\frac{3}{4}x + \frac{43}{4}$

Because N is the point at which the horizontal line (d < 0) touches the circle and is drawn horizontally through K:

Thus, the coordinates can be approximated as:

$N(x_N; y_N) = (2; -3)$

Using the standard form of a circle's equation:

$(x - a)^2 + (y - b)^2 = r^2$

Where (a, b) are the coordinates of N (2, -3) and radius r is the distance from N to T:

Calculating distance gives us radius:

$r = \sqrt{(2 - 0)^2 + (-3 - 0)^2} = \sqrt{4 + 9} = \sqrt{13}$

Substituting into the equation gives:

$(x - 2)^2 + (y + 3)^2 = 13$