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

The equation of the large circle can be determined using its radius, which is equal to the distance OP:

$R = 2\sqrt{13}$

The equation of the circle is given by:

$x^2 + y^2 = R^2$

Substituting for R:

$x^2 + y^2 = (2\sqrt{13})^2 = 52$

Thus, we have:

$x^2 + y^2 = 52$

The smaller circle also has equation of the standard form:

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

Using $M(2, -3)$ as the center, we can denote the radius as (r = \sqrt{13}) (half the diameter PO):

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

Expanding gives:

$x^2 - 4x + 4 + y^2 + 6y + 9 - 13 = 0$

Therefore, $x^2 + y^2 - 4x + 6y = 0$

The line RS is determined by the endpoints, R and S, with R on the positive x-axis and S on the negative x-axis, both equidistant from the center O. Since they are on the circle with radius (R = 2\sqrt{13}):

Coordinates of R are $(2\sqrt{13}, 0)$ and S are $(-2\sqrt{13}, 0)$.

The slope (m) of RS is: $m = \frac{0 - 0}{2\sqrt{13} - (-2\sqrt{13})} = 0$

Thus, the equation of RS is: $y = 0$

To find point N, we reflect point R(2\sqrt{13}, 0) across the y-axis. The coordinates of N become (-2\sqrt{13}, 0).

The length of chord NR can now be obtained using the distance formula:

$d = \text{distance}(N, R) = \sqrt{((-2\sqrt{13}) - (2\sqrt{13}))^2 + (0 - 0)^2} = \sqrt{(-4\sqrt{13})^2} = 4\sqrt{13}$

Reflecting the center M(2, -3) about the x-axis results in point K(2, 3). We need to calculate the length of the common chord of the two circles:

1. Radius of circle at M (since M is the center): (r = \sqrt{13}).
2. Radius of the larger circle: (R = 2\sqrt{13}).

The distance between centers M and K is:

$d = \sqrt{(2 - 2)^2 + (3 - (-3))^2} = \sqrt{0 + 36} = 6$

Using the formula for the length of the common chord:

$L = \sqrt{R^2 - d^2} = \sqrt{(2\sqrt{13})^2 - 6^2} = \sqrt{52 - 36} = \sqrt{16} = 4$

Thus, the length of the common chord is 4.