In Fig. , PQ is tangent at point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.


In the given figure,

In Δ ACO,
OA=OC (Radii of the same circle)
Therefore,
ΔACO is an isosceles triangle.
∠CAB = 30° (Given)
∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)
∠PCO = 90° …(radius is drawn at the point of
contact is perpendicular to the tangent)
Now ∠PCA = ∠PCO – ∠CAO
Therefore,
∠PCA = 90° – 30° = 60°

30599 Views

A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle. 



Radius of the circle, r = 10 cm
Area of sector OPRQ
space equals space 60 to the power of 0 over 360 to the power of 0 space straight x space πr squared
space equals space 1 over 6 space straight x space 3.14 space straight x space left parenthesis 10 right parenthesis squared
space equals 52.33 space cm squared

In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ
space equals fraction numerator square root of 3 over denominator 4 end fraction space straight x space left parenthesis side right parenthesis squared
space equals space fraction numerator square root of 3 over denominator 4 end fraction space straight x space left parenthesis 10 right parenthesis squared
space equals space fraction numerator 100 space square root of 3 over denominator 4 end fraction space cm squared
space equals space 43.30 space cm squared

Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm2
Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
=π(10)2−9.03=314−9.03=304.97 cm2

5151 Views

In fig., l and m are two parallel tangents to a circle with centre O,touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that   DOE = 900


Given:  l and m are two parallel tangents to the circle with centre O touching the circle at A and B respectively. DE is a tangent at the point C, which intersect  l at  D  and  m at  E.To prove:  DOE = 90°Contruction: Join OC.Proof:                     

In ODA and ODC,OA = OC     ( Radii of the same circle )AD = DC      (Length of tangents drawn from an external point to a circle                            are equal )DO = OD       ( Commmon side ) ODA  ODC,         (SSS congruence criterion) DOA = COD          ..........(1)Similarly, OEB OECEOB = COE           ...........(2)Now, AOB is a diameter of the circle. Hence, it is a straight line.DOA +COD + COE + EOB =180°From (1) and (2), we have:2COD + 2COE = 180°COD + COE = 90°DOE = 90°Hence, proved.


Prove that the tangent at any point of a circle is perpendicular to theradius through the point of contact.


Given: A circle with centre O  and a tangent XYto the circle at a point  PTo prove: OP is perpendicular to XY.Construction:Take a point Q on XY other than P and join OQ.

Proof:  Here the point Q must lie outside the circle asif it lies inside the tangent  XY will become secant to the circle.Therefore,  OQ is longer thanthe radius OPof the circle, That is,  OQ> OP.This happens for every point on the line  XY  expect the point  P.So  OP is the shortest of all the distance of the point O to the points on  XY.And hence  OP is  perpendicular to XY.Hence, proved.


Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.


Given: A circle C ( 0, r ) and a tangent l at point A.

To prove: OA  l

                                  

 

Construction: Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.

Proof: We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.

OA = OC  (Radius of the same circle)

Now, OB = OC + BC.

 OB>OC OB>OA OA<OB     ( OC = OA = radius) 

But among all the line segments, joining the point O to a point on AB, the shortest one is the perpendicular from O on AB.

Hence OA is perpendicular to l.

        


Prove that the lengths of tangents drawn from an external point to a circle are equal. 



Proof: We know that a tangent to the circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = ∠OQT = 90°
In ΔOPT and ΔOQT,
OT = OT (Common)
OP = OQ (Radius of the circle)
∠OPT = ∠OQT (90°)
∴ ΔOPT ≅ ΔOQT (RHS congruence criterion)
⇒ TP = TQ (CPCT)
Hence, the lengths of the tangents drawn from an external point to a circle are equal.

1719 Views