(a) The gradient of the tangent through \( (1,5) \) and \( (4,4) \) is

\( \frac{4-5}{4-1}=-\frac13 \).

Using \( (1,5) \),

\( y-5=-\frac13(x-1) \).

So

\( x+3y=16 \).

The other tangent is

\( x-3y=16 \).

Solve simultaneously:

\( x+3y=16 \),

\( x-3y=16 \).

Adding gives \(2x=32\), so \(x=16\).

Then \(y=0\).

Answer: \( (16,0) \).

(b) The tangent through \(A\) has gradient \(-\frac13\), so the normal through \(A\) has gradient \(3\).

The normal also passes through the centre \( (a,0) \).

Therefore

Answer: \( y=3(x-a) \).

(c) Consider the right-angled triangle formed by the centre of the circle, the point of contact, and the point of intersection of the two tangents.

The tangent has gradient \( -\frac13 \), so the radius to the point of contact is perpendicular to the tangent and has gradient \(3\).

Therefore, for the angle between the radius and the horizontal axis,

\[ \tan\theta=3. \]

The radius of the circle is \( \sqrt{40} \). Hence the other perpendicular side of the right-angled triangle is

\[ 3\sqrt{40}. \]

By Pythagoras, the hypotenuse is

\[ \sqrt{(\sqrt{40})^2+(3\sqrt{40})^2} =\sqrt{40+360} =\sqrt{400} =20. \]

This hypotenuse is the distance from the centre \( (a,0) \) to the point of intersection of the tangents \( (16,0) \). Therefore

\[ 16-a=20. \]

So

\[ a=-4. \]

The centre is \( (-4,0) \), and \(r^2=40\). Therefore the equation of the circle is

\[ (x+4)^2+y^2=40. \]

(c) Method(Using FM formula) The centre is \( (a,0) \) and the radius is \( \sqrt{40} \).

The distance from the centre \( (a,0) \) to the tangent \(x+3y=16\) is equal to the radius:

\( \frac{|a-16|}{\sqrt{1^2+3^2}}=\sqrt{40} \).

Since \(a<0\), \( |a-16|=16-a \).

So

\( \frac{16-a}{\sqrt{10}}=\sqrt{40} \).

\( 16-a=\sqrt{400}=20 \).

Thus \(a=-4\).

The centre is \( (-4,0) \) and the radius squared is \(40\).

Answer: \( (x+4)^2+y^2=40 \).