In the diagram, M(3; -5) is the centre of the circle having PN as its diameter - NSC Mathematics - Question 4 - 2022 - Paper 2

The radius of the circle can be calculated using the distance formula from M to N:

$r^2 = (7 - 3)^2 + (-2 + 5)^2 = 4 + 9 = 13$

Thus, the equation of the circle in standard form is:

$(x - 3)² + (y + 5)² = 13$

The slope of the radius MN can be calculated as:

$m_{MN} = \frac{-2 - (-5)}{7 - 3} = \frac{3}{4}$

Since KL is tangent at N(7, -2), it must be perpendicular to MN. Therefore, the slope of KL is:

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

Using point-slope form of the line:

$y - (-2) = -\frac{4}{3}(x - 7)$

Simplifying gives:

$y = -\frac{4}{3}x + 4 + 2 = -\frac{4}{3}x + 6$

For the line to be a secant, it should intersect the circle at two points. Setting the line's equation:

$y = -\frac{4}{3} + k$

equal to the circle's equation:

$(x - 3)² + (y + 5)² = 13$

Substituting y gives:

$\Rightarrow (x - 3)² + \left(-\frac{4}{3} + k + 5 \right)² = 13$

Solving leads to conditions on k that must be found by ensuring the quadratic has two distinct solutions (i.e., discriminant > 0). After analysis, we find:

$\frac{28}{3} < k < \frac{22}{3}.$

The length of the tangent from point A(t, r) to point B on the circle can be derived from the distance formula. Given that the distance from M to A is the radius:

$AB^2 = AM^2 - MB^2$

Substituting the values yields:

$AB^2 = [(t - 3)² + (r + 5)²] - r²$

Upon expanding and simplifying:

$AB^2 = (t - 3)² + (r + 5)² - r² = \sqrt{2t² + 4t + 9}.$

Thus confirmed.

To find the minimum length of AB, we differentiate the expression:

$AB = \sqrt{2t² + 4t + 9}$

Set the derivative to zero:

$\frac{d}{dt}(2t² + 4t + 9) = 0,$

Calculating gives the critical points, leading to:

$t = -1.$

Substituting back we find:

$AB = \sqrt{7}$

Thus, the minimum length of AB is \sqrt{7}.