In the diagram, a circle having centre at the origin passes through P(4; -6) - NSC Mathematics - Question 4 - 2018 - Paper 2

The general equation of a circle with centre (h, k) and radius r is given by:

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

For the large circle whose centre is at the origin (0, 0) and radius is $\sqrt{13}$, the equation becomes:

$x^2 + y^2 = 13$

In standard form:

$x^2 + y^2 - 13 = 0$

The small circle's centre is at M(2, -3) and it has the radius of $r = r_{large} - r_{small}$, where $r_{small} = 0$ (because the tangent point is at O). Thus the radius for this small circle is:

$r_{small} = \sqrt{13}$

The equation of the small circle in standard form is:

$(x - 2)^2 + (y + 3)^2 = 0$

Expanding this gives:

$x^2 - 4x + 4 + y^2 + 6y + 9 = 0$

Combining terms leads to:

$x^2 + y^2 - 4x + 6y + 13 = 0$

The diameter RS is a horizontal line (if we assume RS to be horizontal as is common) passing through the y-coordinate at the origin. Since point R and S both lie on the circle, their coordinates must be found first, which would normally be at O and the corresponding points based on the radius. Hence:

If R and S correspond to (R_x, -3) and (S_x, -3), then the line RS can simply be represented:

$y = mx + c$ where $m=0$ and hence:

$y = -3$

To find the length of chord NR, we first identify point N, which is the reflection of R across the y-axis. If R's coordinates are (x, y), then N will be (-x, y). The distance NR is then calculated using the distance formula:

$NR = \sqrt{((-x - x)^2) + (y - y)^2} = \sqrt{(2x)^2} = 2|x|$

Substituting the appropriate values of x from R will give us the length of NR.

To find the length of the common chord of the two circles, we first need to find their intersection points. Using the equations derived, we equate the two circle equations to find the intersection points.

Then, finding the distance between the points of intersection via the distance formula will yield the required length of the common chord.