Further Mathematics Questions and Answers

The equation for a circle with centre coordinates (a, b) and radius r is

\((x-a)^{2} + (y-b)^{2} = r^{2}\)

Expanding the above equation, we have

\(x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0\) so that

\(x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)

Taking the original equation given, \(3x^{2} + 3y^{2} - 4x + 8y = 2\) and making the coefficients of \(x^{2}\) and \(y^{2}\) = 1,

\(x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}\), comparing, we have

\(2a = \frac{4}{3}; 2b = \frac{-8}{3}\)

\(\implies a = \frac{2}{3}; b = \frac{-4}{3}\)

For equal roots, we have that \(b^{2} = 4ac\), so, given a=1, b = -k and c = 9,

\((-k)^{2} = 4\times1\times9 \implies k^{2} = 36\)

\(k = \sqrt{36} = \pm6\)

\(\frac{1}{(1-\sqrt{3})^{2}}\) 

\((1-\sqrt{3})^{2} = (1-\sqrt{3})(1-\sqrt{3})\)

\(1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}\)

\(\frac{1}{4-2\sqrt{3}}\)

After rationalising (multiplying the denominator and numerator with \(4+2\sqrt{3}\), we have

\(\frac{4+2\sqrt{3}}{4} = 1 + \frac{1}{2}\sqrt{3}\)

\(a \Delta b\) = \(\frac{a+b}{\sqrt{ab}}\)

\(-3\Delta -1\) = \(\frac{-3 + -1}{\sqrt{-3\times -1}}\)

\(\frac{-4}{\sqrt{3}}\), rationalising, we have

\(\frac{-4 \times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} = \frac{-4\sqrt{3}}{3}\)

\(log_{y}\frac{1}{8} = 3 \implies y^{3} = \frac{1}{8}\) (Laws of logarithm)

\(y^{3} = \frac{1}{2^{3}} = (\frac{1}{2})^{3}\)

Equating both sides, we have

\(y = \frac{1}{2}\)