Explanation

a.  \(\frac{\sqrt{75}-3}{\sqrt{3}+1}\)

First, simplify the radical:
\(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\)

So the expression becomes:
\(\frac{5\sqrt{3}-3}{\sqrt{3}+1}\)

Now rationalize the denominator by multiplying top and bottom by \(\sqrt{3}\) - 1):
\(\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\)

Denominator: 3 - 1 = 2
Numerator: 15 - 5\(\sqrt{3} - 3\sqrt{3} + 3 = 18 - 8\sqrt{3}\)

So: \(\frac{18 - 8\sqrt{3}}{2} = 9 - 4\sqrt{3}\).

bi. The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\)  

Substitute the three points:  Point (7, 3): 
\(49 + 9 + 14g + 6f + c = 0 \implies 14g + 6f + c = -58 \quad \text{(1)}\)

Point (2, 8): 
\(4 + 64 + 4g + 16f + c = 0 \implies 4g + 16f + c = - 68 \quad \text{(2)}\)

Point (-3, 3): 
\(9 + 9 - 6g + 6f + c = 0 \implies -6g + 6f + c = -18 \quad \text{(3)}\)

Subtract (2) from (1):  
\(10g - 10f = 10 \implies g - f = 1 \implies g = f + 1 \quad \text{(4)}\)

Subtract (3) from (2):  
\(10g + 10f = -50 \implies g + f = -5 \quad \text{(5)}\)

Solve (4) and (5):  
\((f + 1) + f = -5 \implies 2f + 1 = -5 \implies 2f = -6 \implies f = -3\)
g = -3 + 1 = -2

Substitute \( g = -2 \), \( f = -3 \) into (3):  
\(-6(-2) + 6(-3) + c = -18 \implies 12 - 18 + c = -18 \implies c = -12\)

The equation of the circle is: \(x^2 + y^2 - 4x - 6y - 12 = 0\)

bii. The centre is  -g, -f  = (2, 3).  
The radius \( r \) is given by  \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\)

Radius of the circle = 5 units.

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