### slope, inclination, circle

Question Sample Titled 'slope, inclination, circle'

${P}{\left(-{1},{3}\right)}$ and ${Q}{\left({7},{5}\right)}$ are two points on the circle ${C}$ with centre ${G}$ such that ${P}{Q}$ is a diameter.

 (a) Find the equation of ${C}$. (2 marks) (b) ${L}$ is a straight line passing through ${Q}$ with slope ${m}$ where ${m}\ne\dfrac{{1}}{{4}}$. (i) Express the equation of ${L}$ in terms of ${m}$. (ii) Find the positive value of ${m}$ such that the area of $\triangle{P}{Q}{R}$ is maximum.  (4 marks)

 (a) Centre of ${C}$ $={\left(\dfrac{{-{1}+{7}}}{{2}},\dfrac{{{3}+{5}}}{{2}}\right)}$  $={\left({3},{4}\right)}$ 1A Radius of ${C}$ $=\sqrt{{{\left({3}+{1}\right)}^{{2}}+{\left({4}-{3}\right)}^{{2}}}}$  $=\sqrt{{{17}}}$ Equation of ${C}:$ ${\left({x}-{3}\right)}^{{2}}+{\left({y}-{4}\right)}^{{2}}$ $={17}$ 1A  (bi) Equation of ${L}:$ $\dfrac{{{y}-{5}}}{{{x}-{7}}}$ $={m}$ ${y}-{5}$ $={m}{x}-{7}{m}$ ${m}{x}-{y}-{7}{m}+{5}$ $={0}$ 1A   (bii) Consider ${P}{Q}$ as the base of $\triangle{P}{Q}{R}$. The area is maximum when the height is equal to the radius, resulting ${G}{R}\bot{P}{Q}$ and $\angle{R}{P}{Q}$ $=\angle{R}{Q}{P}={45}^{\circ}$. 1M If ${m}>{0}$, the point ${R}$ will be located as:

${x}$${y}$${P}$${Q}$${R}$${L}$${45}^{\circ}$

 Inclination of ${P}{Q}$ $={{\tan}^{{-{1}}}{\left(\tfrac{{{5}-{3}}}{{{7}+{1}}}\right)}}$ $\therefore{m}$ $={\tan{{\left({{\tan}^{{-{1}}}{\left(\tfrac{{{5}-{3}}}{{{7}+{1}}}\right)}}+{45}^{\circ}\right)}}}$ 1M ${m}$ $={1.67}$ (cor. to 3 sig. fig.) 1A

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