Answer: \((x-2)^2+(y-1)^2=4\), and \(a=-3\pm2\sqrt5\).

(a) The circle is tangent to the vertical lines \(x=0\) and \(x=4\). Therefore its centre is midway between these lines, so the \(x\)-coordinate of the centre is

\(2.\)

The circle is also tangent to the horizontal lines \(y=3\) and \(y=-1\). Therefore the \(y\)-coordinate of the centre is midway between them:

\(1.\)

The distance from the centre \((2,1)\) to each tangent line is \(2\), so the radius is \(2\). Hence the equation of the circle is

\((x-2)^2+(y-1)^2=4.\)

(b) Substitute \(y=2x+a\) into the circle equation:

\((x-2)^2+(2x+a-1)^2=4.\)

Expand:

\(x^2-4x+4+4x^2+4x(a-1)+(a-1)^2=4.\)

Subtract \(4\) from both sides and collect terms:

\(5x^2+4(a-2)x+(a-1)^2=0,\)

as required.

For the line to be tangent to the circle, this quadratic in \(x\) must have exactly one root. Therefore its discriminant is zero:

\([4(a-2)]^2-4(5)(a-1)^2=0.\)

So

\(16(a-2)^2-20(a-1)^2=0.\)

Divide by \(4\):

\(4(a-2)^2-5(a-1)^2=0.\)

Expanding,

\(4(a^2-4a+4)-5(a^2-2a+1)=0.\)

This gives

\(4a^2-16a+16-5a^2+10a-5=0,\)

so

\(-a^2-6a+11=0.\)

Therefore

\(a^2+6a-11=0.\)

Using the quadratic formula,

\(a=\dfrac{-6\pm\sqrt{36+44}}{2}=\dfrac{-6\pm\sqrt{80}}{2}.\)

Hence

\(a=-3\pm2\sqrt5.\)