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

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

If $a = 2$, then: $b = 2 + 1 = 3$

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

To find the radius of the circle, we use the distance formula:

$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting the coordinates of the center $M(2, 3)$ and point $R(6, 0)$:

$r = \sqrt{(6 - 2)^2 + (0 - 3)^2} = \sqrt{4 + 9} = \sqrt{13}$

Thus, the radius is $r = \sqrt{13}$.

Using the distance formula again:

$TR = |6 - 2| = 4$

The length of TR is 4 units.

To determine the equation of the tangent at point $K(5, 7)$, calculate the gradient of the radius to K:

The gradient ($m_{radius}$) = ( \frac{7 - 3}{5 - 2} = \frac{4}{3} )

Therefore, the gradient of the tangent line ($m_{tangent}$) will be the negative reciprocal:

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

Using point-slope form, $y - y_1 = m(x - x_1)$:

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

Thus, the equation is:

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

Since N is a horizontal tangent, its y-coordinate will be the same as K's y-coordinate. Thus, $N(c ; d) = (c, 7)$ where d < 0, we can assume $d = -h$ for some positive $h$. So, taking an example, $N(c ; -2)$ could be valid.

Using the standard circle equation, $(x - a')^2 + (y - b')^2 = r^2$, where the center is N(c, d) and T being (0,0):

The radius (from N to T) is: $r = \sqrt{(0 - c)^2 + (0 - d)^2}$ Substituting into the circle equation: $(x - c)^2 + (y + 2)^2 = r^2$ where you can find $c$ and substitute for complete equation.