Proof geometry

Question Sample Titled 'Proof geometry'

 (a) In Figure (a) , ${D}$ , ${B}$ and ${C}$ are points on circle ${C}_{{1}}$ such that ${D}{C}$ is a diameter. ${B}$ , ${A}$ , ${O}$ and ${C}$ are points on circle ${C}_{{2}}$ . It is given that ${D}{C}$ is the tangent to ${C}_{{2}}$ at ${C}$ , ${D}{B}{A}$ is a straight line, ${B}{A}={O}{A}$, ${E}$ is the point of intersection of ${O}{B}$ and ${A}{C}$ and ${D}{B}\ne{B}{C}$ . Prove that ${D}{C}$$//$${O}{B}$ . (3 marks)

${O}$${D}$${B}$${A}$${C}$${E}$${C}_{{1}}$${C}_{{2}}$Figure (a)

 (b) A rectangular coordinate system, with ${O}$ as the origin, is introduced to figure (a) as shown below. ${O}{A}$ lies on the positive ${x}$-axis and ${O}{C}$ lies on the positive ${y}$-axis. The equation of ${C}_{{2}}$ is ${x}^{{2}}+{y}^{{2}}-{10}{x}-{k}{y}={0}$ and the equation of ${D}{C}$ is ${5}{x}-{4}{y}+{4}{k}={0}$ . Find the value of ${k}$ . (5 marks)

${x}$${y}$${O}$${D}$${B}$${A}$${C}$${E}$${C}_{{1}}$${5}{x}-{4}{y}+{4}{k}={0}$Figure (b)

 (a) Let $\angle{C}{D}{B}$ $=\theta$ . ∵  ${D}{C}$ is a diameter of ${C}_{{1}}$ . ∴  $\angle{D}{B}{C}$ $={90}^{\circ}$ ∠ in semi-circle $\angle{D}{C}{B}$ $={180}^{\circ}-\angle{C}{D}{B}-\angle{D}{B}{C}$ ∠ sum of △  $={180}^{\circ}-\theta-{90}^{\circ}$  $={90}^{\circ}-\theta$  ∵  ${D}{C}$ is the tangent to ${C}_{{2}}$ at ${C}$. $\angle{C}{O}{B}$ $=\angle{D}{C}{B}$ ∠ in alt. segment  $={90}^{\circ}-\theta$  $\angle{C}{O}{A}$ $=\angle{D}{B}{C}$ ext. ∠, cyclic quad.  $={90}^{\circ}$  $\angle{B}{O}{A}$ $=\angle{C}{O}{A}-\angle{C}{O}{B}$  $={90}^{\circ}-{\left({90}^{\circ}-\theta\right)}$  $=\theta$  ∵  ${B}{A}$ $={O}{A}$ ∴  $\angle{O}{B}{A}$ $=\angle{B}{O}{A}=\theta$ base ∠s, isos. △ ∴  $\angle{O}{B}{A}$ $=\angle{C}{D}{B}$ ∴  ${D}{C}$$//$${B}{O}$ corr. ∠s equal  (3 marks)：Correct proof with reasons (2 marks)：Correct proof with missing reason(s) (1 mark)：Incomplete proof with any one correct step and one correct reason

 (b) Equation of ${D}{C}$ : ${5}{x}-{4}{y}+{4}{k}$ $={0}$ ${y}$ $=\dfrac{{5}}{{4}}{x}+{k}$ $\ldots{\left({1}\right)}$ 1M Equation of ${C}_{{2}}$ : ${x}^{{2}}+{y}^{{2}}-{10}{x}-{k}{y}$ $={0}$ $\ldots{\left({2}\right)}$ Put ${\left({1}\right)}$ into ${\left({2}\right)}$ , ${x}^{{2}}+{\left(\dfrac{{5}}{{4}}{x}+{k}\right)}^{{2}}-{10}{x}-{k}{\left(\dfrac{{5}}{{4}}{x}+{k}\right)}$ $={0}$ 1M $\dfrac{{41}}{{16}}{x}^{{2}}+{\left(\dfrac{{5}}{{4}}{k}-{10}\right)}{x}$ $={0}$ ∵  ${D}{C}$ is a tangent to ${C}_{{2}}$ . ∴ $\Delta$ $={0}$ 1M ${\left(\dfrac{{5}}{{4}}{k}-{10}\right)}^{{2}}-{4}\times\dfrac{{41}}{{16}}\times{0}$ $={0}$ 1M ${\left(\dfrac{{5}}{{4}}{k}-{10}\right)}^{{2}}$ $={0}$ ${k}$ $={8}$ 1A

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