Question

Two chords XY and PQ are intersecting at the point A. The line segment Joining X and P is a diameter of the circle, angle XAP= 120° and XY= PQ= 18 cm. Find the distance between the centre of the circle and the point A .Could you please help me?

Answer 1

distance between the center and the point `A=6` cm

Explanation:

Let `O` be the center of the circle.
given `XP` is the diameter, `=>angleXYP=anglePQX=90^@`,
given `angleXAP=120^@, => angleXAQ=60^@`
In `DeltaXAQ, angleQXA=30^@`
`sin30=1/2=(QA)/(XA), => QA:XA=1:2`,
similarly, `YA:PA=1:2`
given `XY=PQ=18` cm
`=> YA=6, AX=12`
as `XA=PA, and O` is midpoint of `XP`,
`=> OA` is perpendicular tp `XP`
As `DeltaOAX and DeltaOAP` are congruent
`=> OA` bisects `angleXAP, => angleXAO=120/2=60^@`
`=> OA=XAcosangleXAO=12cos60=12*1/2=6` cm

Answer 2

`OA=6` cm

Explanation:

Solution 2 (without using trigonometry)
let `O` be the center of the circle,
given `XP` is the diameter, `=> angle XYP=anglePQX=90^@`,
given `angleXAP=120^@, => angleXAQ=anglePAY=60^@`,
`=> angleQXA=30^@`, similarly, `angleYPA=30^@`
As `XQYP` is a cyclic quadrilateral,
`=> anglePQY=anglePXY=a`
as `XY=PQ, and XP` is the common hypotenuse of `DeltaXQP and DeltaPYX, => XQ=PY`
`=> angleXYQ=angleXPQ=anglePQY=anglePXY=a`
`=> angleX+angleY=180^@`,
`=> 30+a+a+90=180^@, => a=30^@`
as `angleAXP=angleAPX, => DeltaAXP` is isosceles,
`=> O` is the midpoint of `XP`,
`=> OA` bisects `angleXAP, => angleXAO=120/2=60^@`
`=> DeltaAOX and DeltaAOP` are congruent,
as `DeltaAOX and DeltaAQX` have equal corresponding angles and the common hypotenuse `AX`,
`=>DeltaAOX and DeltaAQX` are congruent,
`=> color(red)(DeltaAOX, DeltaAOP and DeltaAQX " are congruent")`,
`=> color(red)(AQ=AO, AX=AP, QX=OX=OP)`
let `|QXP|` denote area of `DeltaQXP`,
`=> color(red)(|QXP|)=1/2*QP*QX=1/2*18*QX=color(red)(9QX)`
`color(red)(|QXP|)=|AQX|+|AOX|+|AOP|=3*|AOX|=1/2*3*OA*OX=color(red)(1/2*3*OA*QX)`
`=> |QXP|=9cancel(QX)=3/2*OA*cancel(QX)`
`=> OA=2/3*9=18/3=6` cm