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 can be calculated using the distance formula between M(3, -5) and N(7, -2):

$r² = (7 - 3)² + (-2 + 5)² = 4 + 9 = 13$

The equation of the circle is then:

$(x - 3)² + (y + 5)² = 13$.

Given that KL is tangent to the circle at N(7, -2), we need to find the slope of line KL. The slope of NM, where N is (7,-2) and M is (3,-5), is:

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

Since KL is perpendicular to NM, its slope will be the negative reciprocal:

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

Using point-slope form (y - y₁ = m(x - x₁)), substituting N(7, -2):

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

Thus, the equation becomes:

$y = -\frac{4}{3}x + \frac{28}{3} - 2$

Final simplified form:

$y = -\frac{4}{3}x + \frac{22}{3}$.

To find when the line intersects the circle, we equate:

$\begin{cases} (x - 3)² + (\frac{4}{3}x + k + 5)² = 13 \\ ext{Expand and simplify using standard methods to find k values.} \end{cases}$

ext{Solving leads to a quadratic in }k. Use the discriminant (D) to identify real roots, which implies:}

$D = 0 \text{ or positive indicates intersections - leading ultimately to finding } k.$

Using the formula for the length of a tangent from point A(t, r) to the circle, the length AB is:

$AB² = AM² - MB²$

Where AM = \sqrt{(t - 3)² + (r + 5)²} and MB = 5.

Substituting the values:

$AB² = [(t - 3)² + (r + 5)²] - 25$

This simplifies to yield:

$AB = \sqrt{2t² + 4t + 9}$.

To find the minimum length, we differentiate AB² = 2t² + 4t + 9:

$\frac{d(AB²)}{dt} = 4t + 4$

Setting the derivative to zero to find critical points:

$4t + 4 = 0 \\ t = -1$

Now substituting t back into the length equation gives:

$AB = \sqrt{2(-1)² + 4(-1) + 9} = \sqrt{7}$

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