### Coordinate system of circle involving in-centre and circumcentre and finding possible intersections of circle and tangent

Question Sample Titled 'Coordinate system of circle involving in-centre and circumcentre and finding possible intersections of circle and tangent'

$\triangle{O}{P}{Q}$ is an obtuse-angled triangle. Denote the in-centre and the circumcentre of $\triangle{O}{P}{Q}$ by ${I}$ and ${J}$ respectively. It is given that ${P}$ , ${I}$ and ${J}$ are collinear.

 (a) Prove that ${O}{P}$ $={P}{Q}$ . (3 marks) (b) A rectangular coordinate system is introduced so that the coordinates of ${O}$ and ${Q}$ are ${\left({0},{0}\right)}$ and ${\left({40},{30}\right)}$ respectively while the ${y}$-coordinate of ${P}$ is ${19}$ . Let ${C}$ be the circle which passes through ${O}$ , ${P}$ and ${Q}$ .  (i) Find the equation of ${C}$ . (ii) Let ${L}_{{1}}$ and ${L}_{{2}}$ be two tangents to ${C}$ such that the slope of each tangent is $\dfrac{{3}}{{4}}$ and the ${y}$-intercept of ${L}_{{1}}$ is greater than that of ${L}_{{2}}$ . ${L}_{{1}}$ cuts the ${x}$-axis and the ${y}$-axis at ${S}$ and ${T}$ respectively while ${L}_{{2}}$ cuts the ${x}$-axis and the ${y}$-axis at ${U}$ and ${V}$ respectively. Someone claims that the aera of the trapezium ${S}{T}{U}{V}$ exceeds ${17000}.$ Is the claim correct? Explain your answer.  (9 marks)

${I}$${J}$${P}$${O}$${Q}$${Y}$${Z}$${X}$

 (a) With the notations in the figure, where ${Y}{J}$ and ${Z}{J}$ are perpendicular bisectors of ${O}{P}$ and ${P}{Q}$ respectively. Consider $\triangle{P}{Y}{J}$ and $\triangle{P}{Z}{J}$ , $\angle{Y}{P}{J}$ $=\angle{Z}{P}{J}$ definition of in-centre ${P}{J}$ $={P}{J}$ common side $\angle{P}{J}{Y}$ $={180}^{\circ}-{90}^{\circ}-\angle{Y}{P}{J}$ ∠ sum of △  $={180}^{\circ}-{90}^{\circ}-\angle{Z}{P}{J}$  $=\angle{P}{J}{Z}$ ∴  $\triangle{P}{Y}{J}\stackrel{\sim}{=}\triangle{P}{Z}{J}$ ASA ∴  ${P}{Y}={P}{Z}$ corr. sides, ≅△s ${O}{P}$ $={2}{\left({P}{Y}\right)}$ $={2}{\left({P}{Z}\right)}$ $={P}{Q}$ Marking scheme of (a): Case 1 Any correct proof with correct reasons. 3M Case 2 Any correct proof without reasons. 2M Case 3 Incomplete proof with any one correct step and one correct reason. 1M Caution: Any proof based on the assumption that ${P}{I}{X}{J}$ is collinear should be awarded zero marks.  (b)(i)  Let the coordinates of ${P}$ be ${\left({p},{19}\right)}$ . ${O}{P}$ $={P}{Q}$ proved $\sqrt{{{p}^{{2}}+{19}^{{2}}}}$ $=\sqrt{{{\left({p}-{40}\right)}^{{2}}+{\left({19}-{30}\right)}^{{2}}}}$ 1M for using (a) ${p}^{{2}}+{361}$ $={p}^{{2}}-{80}{p}+{1600}+{121}$ ${80}{p}$ $={1360}$ ${p}$ $={17}$ ∴   The coordinates of ${P}$ is ${\left({17},{19}\right)}$ . 1A Let the equation of ${C}$ be ${x}^{{2}}+{y}^{{2}}+{D}{x}+{E}{y}+{F}={0}$ . 1M for subsituting ${O}$, ${P}$ and ${R}$ Subsitute ${O}{\left({0},{0}\right)}$ into the equation of ${C}$ , ${\left({0}\right)}^{{2}}+{\left({0}\right)}^{{2}}+{D}{\left({0}\right)}+{E}{\left({0}\right)}+{F}$ $={0}$ ${F}$ $={0}$ Subsitute ${Q}{\left({40},{30}\right)}$ and ${P}{\left({17},{19}\right)}$ into the equation of ${C}$ , ${\left({40}\right)}^{{2}}+{\left({30}\right)}^{{2}}+{\left({40}\right)}{D}+{\left({30}\right)}{E}$ $={0}$ ${\left({17}\right)}^{{2}}+{\left({19}\right)}^{{2}}+{\left({17}\right)}{D}+{\left({19}\right)}{E}$ $={0}$ ${4}{D}+{3}{E}+{250}$ $={0}$ $\ldots{\left({1}\right)}$ ${17}{D}+{19}{E}+{650}$ $={0}$ $\ldots{\left({2}\right)}$ Solving, we have ${D}=-{112}$ , ${E}={66}$ . ∴   The equation of ${C}$ is ${x}^{{2}}+{y}^{{2}}-{112}{x}+{66}{y}$ $={0}$ . 1A or equivalent  (b)(ii)  Let the ${y}$-intercepts of ${L}_{{1}}$ and ${L}_{{2}}$ be ${c}$ . ${x}^{{2}}+{y}^{{2}}-{112}{x}+{66}{y}$ $={0}$ ${y}$ $=\dfrac{{3}}{{4}}{x}+{c}$ $\ldots{\left({2}\right)}$ Subsitute ${\left({2}\right)}$ into ${\left({1}\right)}$ : ${x}^{{2}}+{\left(\dfrac{{3}}{{4}}{x}+{c}\right)}^{{2}}-{112}{x}+{66}{\left(\dfrac{{3}}{{4}}{x}+{c}\right)}$ $={0}$ $\dfrac{{25}}{{16}}{x}^{{2}}+{\left(\dfrac{{3}}{{2}}{c}-\dfrac{{125}}{{2}}\right)}{x}+{\left({c}^{{2}}+{66}{c}\right)}$ $={0}$ ${25}{x}^{{2}}+{\left({24}{c}-{1000}\right)}{x}+{\left({16}{c}^{{2}}+{1056}{c}\right)}$ $={0}$ $\ldots{\left(\ast\right)}$ Since ${L}_{{1}}$ and ${L}_{{2}}$ are tangents to ${C}$, ${\left(\ast\right)}$ has only one root. $\Delta$ of ${\left(\ast\right)}$ $={0}$ ${\left({24}{c}-{1000}\right)}^{{2}}-{4}{\left({25}\right)}{\left({16}{c}^{{2}}+{1056}{c}\right)}$ $={0}$ $-{1024}{c}^{{2}}-{153600}{c}+{1000000}$ $={0}$ ${\left({4}{c}+{625}\right)}{\left({4}{c}-{25}\right)}$ $={0}$ ${c}$ $=-\dfrac{{625}}{{4}}$ or ${c}=\dfrac{{25}}{{4}}$ Thus, ${L}_{{1}}:{y}=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}},{L}_{{2}}:{y}=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}$ 2M for both equations of ${L}_{{1}}$ and ${L}_{{2}}$ ∴  ${S}={\left(-\dfrac{{25}}{{3}},{0}\right)},{T}={\left({0},\dfrac{{25}}{{4}}\right)},{U}={\left(\dfrac{{625}}{{3}},{0}\right)},{V}={\left({0},-\dfrac{{625}}{{4}}\right)}$ 1A for both four points The area of trapezium ${S}{T}{U}{V}$ $={(}$area of$\triangle{T}{S}{U}{)}+{(}$area of$\triangle{V}{S}{U}{)}$ accepts other combinations of partitions $=\dfrac{{1}}{{2}}{\left({S}{U}\right)}{\left({O}{T}\right)}+\dfrac{{1}}{{2}}{\left({S}{U}\right)}{\left({O}{V}\right)}$ $=\dfrac{{1}}{{2}}{\left(\dfrac{{625}}{{3}}+\dfrac{{25}}{{3}}\right)}{\left(\dfrac{{25}}{{4}}\right)}+\dfrac{{1}}{{2}}{\left(\dfrac{{625}}{{3}}+\dfrac{{25}}{{3}}\right)}{\left(\dfrac{{625}}{{4}}\right)}$ $={17604}\dfrac{{1}}{{6}}$ r.t. ${17600}$ $\gt{17000}$ Thus, the claim is agreed. 1A The figure below shows the coordinate system. For better illustration purpose, the figure is not drawn to scale.

${x}$${y}$${T}$${S}$${V}$${U}$${O}$${P}$${Q}$${Y}$${L}_{{1}}:$${y}=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}}$${L}_{{2}}:$${y}=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}$

 (a) Note that ${J}$ is the centre of the circumcircle of $\triangle{O}{P}{Q}$ . Join ${J}{O}$ and ${J}{Q}$ . ${J}{O}$ $={J}{P}={J}{Q}$ radii $\angle{J}{P}{O}$ $=\angle{J}{P}{Q}$ definition of in-centre ∴  $\angle{J}{O}{P}$ $=\angle{J}{P}{O}=\angle{J}{P}{Q}=\angle{J}{Q}{P}$ base ∠s, isos. △  Consider $\triangle{J}{O}{P}$ and $\triangle{J}{Q}{P}$ , $\angle{J}{O}{P}$ $=\angle{J}{P}{Q}$ proved $\angle{J}{P}{O}$ $=\angle{J}{P}{Q}$ proved ${P}{J}$ $={P}{J}$ common ∴  $\triangle{J}{O}{P}$ $\stackrel{\sim}{=}\triangle{J}{Q}{P}$ AAS ∴  ${O}{P}$ $={P}{Q}$ corr. sides, ≅△s  Marking scheme of (a): Case 1 Any correct proof with correct reasons. 3M Case 2 Any correct proof without reasons. 2M Case 3 Incomplete proof with any one correct step and one correct reason. 1M Caution: Any proof based on the assumption that ${P}{I}{X}{J}$ is collinear should be awarded zero marks.  (b)(i)  Let the coordinates of ${P}$ be ${\left({p},{19}\right)}$ . ${O}{P}$ $={P}{Q}$ proved $\sqrt{{{p}^{{2}}+{19}^{{2}}}}$ $=\sqrt{{{\left({p}-{40}\right)}^{{2}}+{\left({19}-{30}\right)}^{{2}}}}$ 1M for using (a) ${p}^{{2}}+{361}$ $={p}^{{2}}-{80}{p}+{1600}+{121}$ ${80}{p}$ $={1360}$ ${p}$ $={17}$ $\therefore$The coordinates of ${P}$ is ${\left({17},{19}\right)}$ . 1A  Denote ${K}$ be the centre of circle ${C}$ . Denote the perpendicular bisector of ${O}{P}$ and ${P}{Q}$ be ${P}_{{1}}$ and ${P}_{{2}}$ respectively. Since ${P}_{{1}}\bot{O}{P}$ and ${P}_{{2}}\bot{P}{Q}$ , they must both pass through ${K}$ . line _|_ chord and bisect chord passes through center Solving the system of simultaneous equations below would produce the intersection, i.e. coordinates of ${K}$ . ${P}_{{1}}$ ${P}_{{2}}$ ${\left({x}-{0}\right)}^{{2}}+{\left({y}-{0}\right)}^{{2}}$ $={\left({x}-{17}\right)}^{{2}}+{\left({y}-{19}\right)}^{{2}}$ 1M for correct method attempting to find center of ${C}$ ${\left({x}-{17}\right)}^{{2}}+{\left({y}-{19}\right)}^{{2}}$ $={\left({x}-{40}\right)}^{{2}}+{\left({y}-{30}\right)}^{{2}}$ ${17}{x}+{19}{y}-{325}$ $={0}$ $\ldots{\left({1}\right)}$ ${23}{x}+{11}{y}-{925}$ $={0}$ $\ldots{\left({2}\right)}$ Solving, we have ${\left({56},-{33}\right)}$ . ∴   The coordinates of ${K}$ are ${\left({56},-{33}\right)}$ . The equation of ${C}$ is  ${\left({x}-{56}\right)}^{{2}}+{\left({y}+{33}\right)}^{{2}}$ $={\left(\sqrt{{{56}^{{2}}+{\left(-{33}\right)}^{{2}}}}\right)}^{{2}}$ ${\left({x}-{56}\right)}^{{2}}+{\left({y}+{33}\right)}^{{2}}$ $={4225}$ 1A or equivalent  (b)(ii)  ${K}={\left({56},-{33}\right)}$ by (b)(i) The radius of ${C}=\sqrt{{{56}^{{2}}+{\left(-{33}\right)}^{{2}}}}={65}$ Notice that the slope of ${O}{Q}$ is equivalent to that of ${L}_{{1}}$ and ${L}_{{2}}$ . Thus, ${L}_{{1}}$ is tangential to ${C}$ at ${P}$ . The equation of ${L}_{{1}}$ is ${y}-{19}$ $=\dfrac{{3}}{{4}}{\left({x}-{17}\right)}$ ${y}$ $=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}}$  Denote ${Y}{\left({a},{b}\right)}$ be a point on ${L}_{{2}}$ such that ${P}{Y}$ is a diameter to circle ${C}$ . $\dfrac{{{17}+{a}}}{{2}}={56}$ and $\dfrac{{{19}+{b}}}{{2}}=-{33}$ mid-pt. theorem ${a}={95}$ and ${b}=-{33}$ The equation of ${L}_{{2}}$ is ${y}-{\left(-{33}\right)}$ $=\dfrac{{3}}{{4}}{\left({x}-{95}\right)}$ ${y}$ $=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}$ 2M for both equations of ${L}_{{1}}$ and ${L}_{{2}}$ ∴  ${S}={\left(-\dfrac{{25}}{{3}},{0}\right)},{T}={\left({0},\dfrac{{25}}{{4}}\right)},{U}={\left(\dfrac{{625}}{{3}},{0}\right)},{V}={\left({0},-\dfrac{{625}}{{4}}\right)}$ 1A for both four points The area of trapezium ${S}{T}{U}{V}$ $=\dfrac{{{S}{V}}}{{2}}{\left({S}{T}+{V}{U}\right)}$ $={\left(\dfrac{{{65}+{65}}}{{2}}\right)}{\left(\sqrt{{{\left(-\dfrac{{25}}{{3}}\right)}^{{2}}+{\left(\dfrac{{25}}{{4}}\right)}^{{2}}}}+\sqrt{{{\left(\dfrac{{625}}{{3}}\right)}^{{2}}+{\left(-\dfrac{{625}}{{4}}\right)}^{{2}}}}\right)}$ ${S}{V}$=diameter of ${C}$ 1A f.t. $={17604}\dfrac{{1}}{{6}}$ r.t. ${17600}$ $\gt{17000}$ Thus, the claim is agreed. 1A  The figure below shows the coordinate system. For better illustration purpose, the figure is not drawn to scale.

${I}$${J}$${P}$${O}$${Q}$${Y}$${Z}$${X}$
${x}$${y}$${T}$${S}$${V}$${U}$${O}$${P}$${Q}$${Y}$${L}_{{1}}:$${y}=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}}$${L}_{{2}}:$${y}=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}$${K}$

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