(a) Simplify \(\frac{\sqrt{75} - 3}{\sqrt{3} + 1}\), leaving your answer in the form \(a + b\sqrt{c}\); where a, b and c are rational numbers. (b) The point...

Answer Details

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

Since \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}\), the expression is \(\dfrac{5\sqrt{3}-3}{\sqrt{3}+1}\). Rationalise by multiplying by \(\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\):

\[\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{15-5\sqrt{3}-3\sqrt{3}+3}{3-1}=\frac{18-8\sqrt{3}}{2}=9-4\sqrt{3}.\]

So \(a=9,\ b=-4,\ c=3\).

(b) Let the circle be \(x^2+y^2+2gx+2fy+c=0\).

\((7,3):\ 58+14g+6f+c=0\)  (i)
\((-3,3):\ 18-6g+6f+c=0\)  (ii)
\((2,8):\ 68+4g+16f+c=0\)  (iii)

(i) - (ii): \(40+20g=0\Rightarrow g=-2\).

(ii): \(18-6(-2)+6f+c=0\Rightarrow 6f+c=-30\)  (iv).
(iii): \(68+4(-2)+16f+c=0\Rightarrow 16f+c=-60\)  (v).

(v) - (iv): \(10f=-30\Rightarrow f=-3\), then \(c=-30-6(-3)=-12\).

(i) Equation: \(x^2+y^2-4x-6y-12=0\), i.e. \((x-2)^2+(y-3)^2=25\).

(ii) Radius: centre \((-g,-f)=(2,3)\), \(r=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=\textbf{5}\).