Answer: (a) \(3x-4y=17\). (b) The circles do not intersect.

(a) The centre of the circle is \((5,12)\).

The gradient of the radius from \((5,12)\) to \((11,4)\) is

\(\frac{4-12}{11-5}=-\frac43.\)

The tangent is perpendicular to the radius, so its gradient is \(\frac34\).

Using the point \((11,4)\),

\(y-4=\frac34(x-11).\)

Multiplying by \(4\),

\(4y-16=3x-33.\)

Therefore

\(3x-4y=17.\)

(b) The centre of \(C\) is \((5,12)\) and its radius is \(10\).

The circle \(x^2+y^2=4\) has centre \((0,0)\) and radius \(2\).

The distance between the centres is

\(\sqrt{5^2+12^2}=13.\)

The sum of the radii is \(10+2=12\). Since \(13\gt 12\), the circles are separate and do not intersect.