Aieee. I have many questions...

1a) Ah! I think it was a careless mistake on my part. Thank you (tkhunny and mr fantastic) for correcting me!

b) Ermm... may I know how did both of you come to an answer? Was it by trial and error?

Mr F says: More or less (for myself, anyway) - just playing around drawing circles.
2) I understand that I need to find a tangent to the y = tan x graph for

(pi/2) < x < (3pi/2), but why is the general equation of passing through the two points of the circle tangent to the tan graph y= (-b/a)x? Why does it have a negative gradient?

Mr F says: Look at the attachment and you'll clearly see why the gradient is negative.
If the points that the tangent touch are (a,b) and (-a, -b), shouldn't the gradient be positive?

Mr F says: No. But the reason is not clear from my poor choice of notation for the points. Look at the attachment.
Just to confirm, there is no constant term in the equation because the line passes through the origin so the origin is the y-intercept right?

Mr F says: Correct. You can justify this either by appealing to symmetry, or deriving the equation from the two tangent points (-a, b) and (a, -b):

$\displaystyle m = \frac{-b - b}{a - (-a)} = -\frac{b}{a}$.

Then using the model $\displaystyle y - y_1 = m(x - x_1) \, $:

$\displaystyle y - (-b) = -\frac{b}{a} (x - a) \Rightarrow y + b = -\frac{b}{a}x + b \Rightarrow y = -\frac{b}{a}$.
Just to confirm again, the general rule for the radius is because of Pythagoras' Theorem right?

Mr F says: Yes.
What is a transcendental equation?

Mr F says: It's an equation involving 'mixed' functions eg. trig and polynomial, exponential and polynomial, trig and exponential etc.
Why do you take the largest negative solution? Why not a positive solution??

Mr F says: Gets back to my poor choice of notation ..... If you look at the attachment, you'll now see that the smallest postive value of a is required.
If R = 2.55702, the equation of the circle is: y = (2.55^2 - x^2)^(1/2). [R rounded off to 3 significant figures.]

Mr F says: Almost. You need a $\displaystyle \pm$ in front .....
Thank you!

Mr F says: You're welcome.