Since the x-axis is tangent to C1, the centre of the circle is 13 units vertically above or below the x-axis.

y-coordinate of centre is either 13 or - 13.

Let equation of C1 be

(x - a)² + (y - b)² = r² , whereby the coordinates of the centre are (a,b)

We know that
r = 13,
b = 13 or -13
and the circle passes through the point (10,18)


When b = 13, x = 10, y = 18, r = 13,

(10 - a)² + (18 - 13)² = 13²
(10 - a)² + 25 = 169
(10 - a)² = 144
10 - a = ±√144

10 - a = 12 or a = -12
a = 10 - 12 or 10 + 12
a = -2 or a = 22


When b = -13,

(10 - a)² + (18 - (-13))² = 13²
(10 - a)² + 961 = 169
(10 - a)² = -792

Since (10 - a)² ≥ 0 for all real a, there are no real solutions/roots for this equation.


Therefore the possible equations of C1 are:


(x + 2)² + (y - 13)² = 169

Or

(x - 22)² + (y - 13)² = 169

I doubt that the b = -13 case can be accepted here, since the presence of a point P above the x-axis implies that the x-axis must be the tangent which lies fully below the circle, so that the circle (and therefore its centre as well) will lie fully above the x-axis.

b = -13 is based on that piece of information alone. If you pair it with point P, you'll realise that there is a complex solution as well.

The working is to show that for b = -13, there are no real roots.

Some students will not realise that b can only be 13 because they didn't put the two pieces of information together. This working allows them to stop in their tracks and eventually realise that

If one were to justify why b = 13, it should go like this :

① Since the x-axis is tangent to the circle, it touches the circle at only 1 point.

So except that point, the circle is either lies fully above or fully below the x-axis.


② Since the circle passes through P(10,18), and P is above the x-axis,

the circle has to be above the x-axis to pass through it.

③ Since tangent is perpendicular to radius, and the radius is 13 units,

the centre of the circle is 13 units directly/vertically above the x-axis (where y = 0)

Therefore the y-coordinate = 13