Two circles s and c touch internally at B, as shown - Leaving Cert Mathematics - Question 4 - 2015

To find the radius of the circle s, we start from the equation:

$(x - 1)^2 + (y + 6)^2 = 360.$
The radius $r$ is therefore: $r = \sqrt{360} = \sqrt{36 \times 10} = 6\sqrt{10}.$ Thus, $a = 6$.

Given the coordinates of B (7, 12) and that the radius of circle c is one-third the radius of circle s:

1. The radius of s is $6\sqrt{10}$, so the radius of c is: $r_c = \frac{1}{3} \times 6\sqrt{10} = 2\sqrt{10}.$
2. Since K lies internally from B towards the center of s, we find: $|AK| : |KB| = 1 : 2.$ Using the section formula in the ratio of 1:2, we calculate: $K = \left(\frac{2 \times 7 + 1 \times 1}{2 + 1}, \frac{2 \times 12 + 1 \times (-6)}{2 + 1}\right) = (5, 6).$

The equation for circle c can be derived using its center K(5, 6) and radius $2\sqrt{10}$:

$(x - 5)^2 + (y - 6)^2 = (2\sqrt{10})^2 = 40.$ Thus, the equation of c is: $(x - 5)^2 + (y - 6)^2 = 40.$

To find the equation of the common tangent:

1. First, calculate the slope of line AB: $\text{slope of } AB = \frac{12 - 6}{7 - 1} = 1.$
2. The slope of the tangent, being perpendicular to AB, is: $\text{slope of tangent} = -1.$
3. Using point-slope form to find the equation of the tangent at B (7, 12):

ightarrow y = -x + 19.$$ 4. Converting to the required form: $$x + y - 19 = 0.$$