Good morning Mirabelle!

To do this question, we must make use of the fact that the perpendicular bisector of any chord in a circle will pass through the centre of the said circle.

A chord simply means a straight line joining two random points lying along the circumference of the circle.

A perpendicular bisector means a line which cuts another line segment, at right angles, into two equal halves.

Here, I will use the coordinates A (-3, 2) and B (13, 2) along the circumference of the circle to form a chord AB. This is a horizontal chord, since the y-coordinates of both A and B are equal at 2.

The perpendicular bisector of this chord AB must therefore be a vertical line. Since the perpendicular bisector cuts the chord into two equal halves, the bisector must pass through the midpoint of AB, which is (5, 2).

So, the equation of the perpendicular bisector of AB is x = 5.

The centre of the circle must lie along the line x = 5, and therefore the x-coordinate of the centre of the circle must be 5.

Since x = -5 is a tangent to the circle. the point on the circle where x = -5 must be the leftmost point of the circle. This point is directly to the left of the centre of the circle.

So, radius of circle
= x-coordinate of centre - x-coordinate of leftmost point
= 5 - (-5)
= 10

Suppose the centre of the circle is called (5, y).

Length of the line joining (5, y) and (13, 2)
= sqrt [(13 - 5)^2 + (2 - y)^2]
= sqrt [64 + (2 - y)^2]

Since this length is also a radius of the circle

sqrt [64 + (2 - y)^2] = 10
64 + (2 - y)^2 = 100
(2 - y)^2 = 36
2 - y = 6 or -6
y = -4 or 8

Since the centre of the circle lies above the x-axis, the y-coordinate of the centre of the circle must be positive.

Therefore, we accept the value y = 8.

The centre of the circle is (5. 8).

The equation of the circle is given by
(x - 5)² + (y - 8)² = 10²
x² - 10x + 25 + y² - 16y + 64 = 100
x² + y² - 10x - 16y - 11 = 0

Of course, the tangents which are parallel to the x-axis (i.e. the horizontal tangents) must touch the uppermost and bottommost points of the circle.

Since the centre of the circle is (5, 8) and the radius of the circle is 10 units,

uppermost point = 5 + 10 = 15
bottommost point = 5 - 10 = -5

So, the equations of the two tangents to the circle are y = 15 and y = -5.

Remember that the perpendicular bisector of a chord will pass through the midpoint of that chord.

So, the point (8, 13) must lie along the perpendicular bisector and its related chord.

Gradient of line joining (5, 8) and (8, 13)
= (13 - 8) / (8 - 5)
= 5/3

Since this perpendicular bisector is perpendicular to the chord FG, gradient of FG = - 3/5.

The chord also contains the point (8, 13).

y = 3/5 x + c
13 = -3/5 (8) + c
13 = -24/5 + c
89/5 = c

y = 3/5 x + 89/5
5y = 3x + 89

Thank you so much!!

Mirabelle, a small correction:

y = -3/5 x + 89/5
5y = -3x + 89