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'

題目

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

(a)Prove that OP{O}{P}=PQ={P}{Q} .(3 marks)
(b)A rectangular coordinate system is introduced so that the coordinates of O{O} and Q{Q} are (0,0){\left({0},{0}\right)} and (40,30){\left({40},{30}\right)} respectively while the y{y}-coordinate of P{P} is 19{19} . Let C{C} be the circle which passes through O{O} , P{P} and Q{Q} .
(i)Find the equation of C{C} .
(ii)Let L1{L}_{{1}} and L2{L}_{{2}} be two tangents to C{C} such that the slope of each tangent is 34\dfrac{{3}}{{4}} and the y{y}-intercept of L1{L}_{{1}} is greater than that of L2{L}_{{2}} . L1{L}_{{1}} cuts the x{x}-axis and the y{y}-axis at S{S} and T{T} respectively while L2{L}_{{2}} cuts the x{x}-axis and the y{y}-axis at U{U} and V{V} respectively. Someone claims that the aera of the trapezium STUV{S}{T}{U}{V} exceeds 17000.{17000}. Is the claim correct? Explain your answer.
(9 marks)

題解

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

(a)With the notations in the figure, where YJ{Y}{J} and ZJ{Z}{J} are perpendicular bisectors of OP{O}{P} and PQ{P}{Q} respectively.
Consider △PYJ\triangle{P}{Y}{J} and △PZJ\triangle{P}{Z}{J} ,
∠YPJ\angle{Y}{P}{J}=∠ZPJ=\angle{Z}{P}{J}definition of in-centre
PJ{P}{J}=PJ={P}{J}common side
∠PJY\angle{P}{J}{Y}=180∘−90∘−∠YPJ={180}^{\circ}-{90}^{\circ}-\angle{Y}{P}{J}∠ sum of △
=180∘−90∘−∠ZPJ={180}^{\circ}-{90}^{\circ}-\angle{Z}{P}{J}
=∠PJZ=\angle{P}{J}{Z}
∴  △PYJ=∼△PZJ\triangle{P}{Y}{J}\stackrel{\sim}{=}\triangle{P}{Z}{J}ASA
∴  PY=PZ{P}{Y}={P}{Z}corr. sides, ≅△s
OP{O}{P}
=2(PY)={2}{\left({P}{Y}\right)}
=2(PZ)={2}{\left({P}{Z}\right)}
=PQ={P}{Q}
Marking scheme of (a):
Case 1Any correct proof with correct reasons.3M
Case 2Any correct proof without reasons.2M
Case 3Incomplete proof with any one correct step and one correct reason.1M
Caution: Any proof based on the assumption that PIXJ{P}{I}{X}{J} is collinear should be awarded zero marks.
(b)(i)
Let the coordinates of P{P} be (p,19){\left({p},{19}\right)} .
OP{O}{P}=PQ={P}{Q}proved
p2+192\sqrt{{{p}^{{2}}+{19}^{{2}}}}=(p−40)2+(19−30)2=\sqrt{{{\left({p}-{40}\right)}^{{2}}+{\left({19}-{30}\right)}^{{2}}}}1Mfor using (a)
p2+361{p}^{{2}}+{361}=p2−80p+1600+121={p}^{{2}}-{80}{p}+{1600}+{121}
80p{80}{p}=1360={1360}
p{p}=17={17}
∴   The coordinates of P{P} is (17,19){\left({17},{19}\right)} .1A
Let the equation of C{C} be x2+y2+Dx+Ey+F=0{x}^{{2}}+{y}^{{2}}+{D}{x}+{E}{y}+{F}={0} .1Mfor subsituting O{O}, P{P} and R{R}
Subsitute O(0,0){O}{\left({0},{0}\right)} into the equation of C{C} ,
(0)2+(0)2+D(0)+E(0)+F{\left({0}\right)}^{{2}}+{\left({0}\right)}^{{2}}+{D}{\left({0}\right)}+{E}{\left({0}\right)}+{F}=0={0}
F{F}=0={0}
Subsitute Q(40,30){Q}{\left({40},{30}\right)} and P(17,19){P}{\left({17},{19}\right)} into the equation of C{C} ,
(40)2+(30)2+(40)D+(30)E{\left({40}\right)}^{{2}}+{\left({30}\right)}^{{2}}+{\left({40}\right)}{D}+{\left({30}\right)}{E}=0={0}
(17)2+(19)2+(17)D+(19)E{\left({17}\right)}^{{2}}+{\left({19}\right)}^{{2}}+{\left({17}\right)}{D}+{\left({19}\right)}{E}=0={0}
4D+3E+250{4}{D}+{3}{E}+{250}=0={0}…(1)\ldots{\left({1}\right)}
17D+19E+650{17}{D}+{19}{E}+{650}=0={0}…(2)\ldots{\left({2}\right)}
Solving, we have D=−112{D}=-{112} , E=66{E}={66} .
∴   The equation of C{C} is x2+y2−112x+66y{x}^{{2}}+{y}^{{2}}-{112}{x}+{66}{y}=0={0} .1Aor equivalent
(b)(ii)
Let the y{y}-intercepts of L1{L}_{{1}} and L2{L}_{{2}} be c{c} .
x2+y2−112x+66y{x}^{{2}}+{y}^{{2}}-{112}{x}+{66}{y}=0={0}
y{y}=34x+c=\dfrac{{3}}{{4}}{x}+{c}…(2)\ldots{\left({2}\right)}
Subsitute (2){\left({2}\right)} into (1){\left({1}\right)} :
x2+(34x+c)2−112x+66(34x+c){x}^{{2}}+{\left(\dfrac{{3}}{{4}}{x}+{c}\right)}^{{2}}-{112}{x}+{66}{\left(\dfrac{{3}}{{4}}{x}+{c}\right)}=0={0}
2516x2+(32c−1252)x+(c2+66c)\dfrac{{25}}{{16}}{x}^{{2}}+{\left(\dfrac{{3}}{{2}}{c}-\dfrac{{125}}{{2}}\right)}{x}+{\left({c}^{{2}}+{66}{c}\right)}=0={0}
25x2+(24c−1000)x+(16c2+1056c){25}{x}^{{2}}+{\left({24}{c}-{1000}\right)}{x}+{\left({16}{c}^{{2}}+{1056}{c}\right)}=0={0}…(∗)\ldots{\left(\ast\right)}
Since L1{L}_{{1}} and L2{L}_{{2}} are tangents to C{C}, (∗){\left(\ast\right)} has only one root.
Δ\Delta of (∗){\left(\ast\right)}=0={0}
(24c−1000)2−4(25)(16c2+1056c){\left({24}{c}-{1000}\right)}^{{2}}-{4}{\left({25}\right)}{\left({16}{c}^{{2}}+{1056}{c}\right)}=0={0}
−1024c2−153600c+1000000-{1024}{c}^{{2}}-{153600}{c}+{1000000}=0={0}
(4c+625)(4c−25){\left({4}{c}+{625}\right)}{\left({4}{c}-{25}\right)}=0={0}
c{c}=−6254=-\dfrac{{625}}{{4}} or c=254{c}=\dfrac{{25}}{{4}}
Thus, L1:y=34x+254,L2:y=34x−6254{L}_{{1}}:{y}=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}},{L}_{{2}}:{y}=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}2Mfor both equations of L1{L}_{{1}} and L2{L}_{{2}}
∴  S=(−253,0),T=(0,254),U=(6253,0),V=(0,−6254){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)}1Afor both four points
The area of trapezium STUV{S}{T}{U}{V}
=(={(}area of△TSU)+(\triangle{T}{S}{U}{)}+{(}area of△VSU)\triangle{V}{S}{U}{)}accepts other combinations of partitions
=12(SU)(OT)+12(SU)(OV)=\dfrac{{1}}{{2}}{\left({S}{U}\right)}{\left({O}{T}\right)}+\dfrac{{1}}{{2}}{\left({S}{U}\right)}{\left({O}{V}\right)}
=12(6253+253)(254)+12(6253+253)(6254)=\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)}
=1760416={17604}\dfrac{{1}}{{6}}r.t. 17600{17600}
>17000\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{x}y{y}T{T}S{S}V{V}U{U}O{O}P{P}Q{Q}Y{Y}L1:{L}_{{1}}:y=34x+254{y}=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}}L2:{L}_{{2}}:y=34x−6254{y}=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}
其他方法

(a)Note that J{J} is the centre of the circumcircle of △OPQ\triangle{O}{P}{Q} .
Join JO{J}{O} and JQ{J}{Q} .
JO{J}{O}=JP=JQ={J}{P}={J}{Q}radii
∠JPO\angle{J}{P}{O}=∠JPQ=\angle{J}{P}{Q}definition of in-centre
∴  ∠JOP\angle{J}{O}{P}=∠JPO=∠JPQ=∠JQP=\angle{J}{P}{O}=\angle{J}{P}{Q}=\angle{J}{Q}{P}base ∠s, isos. △
Consider △JOP\triangle{J}{O}{P} and △JQP\triangle{J}{Q}{P} ,
∠JOP\angle{J}{O}{P}=∠JPQ=\angle{J}{P}{Q}proved
∠JPO\angle{J}{P}{O}=∠JPQ=\angle{J}{P}{Q}proved
PJ{P}{J}=PJ={P}{J}common
∴  △JOP\triangle{J}{O}{P}=∼△JQP\stackrel{\sim}{=}\triangle{J}{Q}{P}AAS
∴  OP{O}{P}=PQ={P}{Q}corr. sides, ≅△s
Marking scheme of (a):
Case 1Any correct proof with correct reasons.3M
Case 2Any correct proof without reasons.2M
Case 3Incomplete proof with any one correct step and one correct reason.1M
Caution: Any proof based on the assumption that PIXJ{P}{I}{X}{J} is collinear should be awarded zero marks.
(b)(i)
Let the coordinates of P{P} be (p,19){\left({p},{19}\right)} .
OP{O}{P}=PQ={P}{Q}proved
p2+192\sqrt{{{p}^{{2}}+{19}^{{2}}}}=(p−40)2+(19−30)2=\sqrt{{{\left({p}-{40}\right)}^{{2}}+{\left({19}-{30}\right)}^{{2}}}}1Mfor using (a)
p2+361{p}^{{2}}+{361}=p2−80p+1600+121={p}^{{2}}-{80}{p}+{1600}+{121}
80p{80}{p}=1360={1360}
p{p}=17={17}
∴\thereforeThe coordinates of P{P} is (17,19){\left({17},{19}\right)} .1A
Denote K{K} be the centre of circle C{C} .
Denote the perpendicular bisector of OP{O}{P} and PQ{P}{Q} be P1{P}_{{1}} and P2{P}_{{2}} respectively.
Since P1⊥OP{P}_{{1}}\bot{O}{P} and P2⊥PQ{P}_{{2}}\bot{P}{Q} , they must both pass through K{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{K} .
P1{P}_{{1}}
P2{P}_{{2}}
(x−0)2+(y−0)2{\left({x}-{0}\right)}^{{2}}+{\left({y}-{0}\right)}^{{2}}=(x−17)2+(y−19)2={\left({x}-{17}\right)}^{{2}}+{\left({y}-{19}\right)}^{{2}}1Mfor correct method attempting to find center of C{C}
(x−17)2+(y−19)2{\left({x}-{17}\right)}^{{2}}+{\left({y}-{19}\right)}^{{2}}=(x−40)2+(y−30)2={\left({x}-{40}\right)}^{{2}}+{\left({y}-{30}\right)}^{{2}}
17x+19y−325{17}{x}+{19}{y}-{325}=0={0}…(1)\ldots{\left({1}\right)}
23x+11y−925{23}{x}+{11}{y}-{925}=0={0}…(2)\ldots{\left({2}\right)}
Solving, we have (56,−33){\left({56},-{33}\right)} .
∴   The coordinates of K{K} are (56,−33){\left({56},-{33}\right)} .
The equation of C{C} is
(x−56)2+(y+33)2{\left({x}-{56}\right)}^{{2}}+{\left({y}+{33}\right)}^{{2}}=(562+(−33)2)2={\left(\sqrt{{{56}^{{2}}+{\left(-{33}\right)}^{{2}}}}\right)}^{{2}}
(x−56)2+(y+33)2{\left({x}-{56}\right)}^{{2}}+{\left({y}+{33}\right)}^{{2}}=4225={4225}1Aor equivalent
(b)(ii)
K=(56,−33){K}={\left({56},-{33}\right)}by (b)(i)
The radius of C=562+(−33)2=65{C}=\sqrt{{{56}^{{2}}+{\left(-{33}\right)}^{{2}}}}={65}
Notice that the slope of OQ{O}{Q} is equivalent to that of L1{L}_{{1}} and L2{L}_{{2}} .
Thus, L1{L}_{{1}} is tangential to C{C} at P{P} .
The equation of L1{L}_{{1}} is
y−19{y}-{19}=34(x−17)=\dfrac{{3}}{{4}}{\left({x}-{17}\right)}
y{y}=34x+254=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}}
Denote Y(a,b){Y}{\left({a},{b}\right)} be a point on L2{L}_{{2}} such that PY{P}{Y} is a diameter to circle C{C} .
17+a2=56\dfrac{{{17}+{a}}}{{2}}={56} and 19+b2=−33\dfrac{{{19}+{b}}}{{2}}=-{33}mid-pt. theorem
a=95{a}={95} and b=−33{b}=-{33}
The equation of L2{L}_{{2}} is
y−(−33){y}-{\left(-{33}\right)}=34(x−95)=\dfrac{{3}}{{4}}{\left({x}-{95}\right)}
y{y}=34x−6254=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}2Mfor both equations of L1{L}_{{1}} and L2{L}_{{2}}
∴  S=(−253,0),T=(0,254),U=(6253,0),V=(0,−6254){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)}1Afor both four points
The area of trapezium STUV{S}{T}{U}{V}
=SV2(ST+VU)=\dfrac{{{S}{V}}}{{2}}{\left({S}{T}+{V}{U}\right)}
=(65+652)((−253)2+(254)2+(6253)2+(−6254)2)={\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)}SV{S}{V}=diameter of C{C}1Af.t.
=1760416={17604}\dfrac{{1}}{{6}}r.t. 17600{17600}
>17000\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{I}J{J}P{P}O{O}Q{Q}Y{Y}Z{Z}X{X}
x{x}y{y}T{T}S{S}V{V}U{U}O{O}P{P}Q{Q}Y{Y}L1:{L}_{{1}}:y=34x+254{y}=\dfrac{{3}}{{4}}{x}+\dfrac{{25}}{{4}}L2:{L}_{{2}}:y=34x−6254{y}=\dfrac{{3}}{{4}}{x}-\dfrac{{625}}{{4}}K{K}

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