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A circle is a two-dimensional geometrical figure formed by a set of all points at equidistant from a fixed point on the same plane. The circle has both area and perimeter. The perimeter of the circle is also called the circumference of a circle and an area of a circle is the region circumscribed by it.
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Keyterms: Circle, Area, Perimeter, Plane, Circular region, Radius, Diameter, Circumference, Chord, Segments, Sector
Circle
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The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a Circle. The fixed point is called the centre of the circle.
Circle
A circle divides the plane on which it lies into three parts. They are: Inside the circle which is also called the interior of the circle; the circle and outside the circle, which is also called the exterior of the circle. The circle and its interior forms the circular region.
There can be three possibilities for a circle and a line on a plane.
They are: non-intersecting (i), a single common point (ii) and two common points (iii).
Radius: The distance from any point on the circle to a fixed point is called radius.
Diameter: The diameter is twice the radius. All the diameters have the same length.
Circumference: Circumference is the length of the complete circle. It is also called the perimeter of the circle.
Chord: Any line segment made by joining any two points on the boundary of the circle is the chord of the circle. The longest chord on the circle which passes through the centre is Diameter.
Segment: The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle. There are two types of segments, which are major and minor segments.
Sector: The region between an arc and the two radii, joining the centre to the end points of the arc is called sector. There are two types of sectors, named major and minor sectors.
Secant: When a line intersects a circle in such a way that there are two common points then that line is called secant.
Tangent to a Circle: A tangent to a circle is a line that intersects the circle at only one point. The tangent to a circle is a special case of secant, when the two end points of its corresponding chord coincide.
There could be only one tangent at a point of contact. Tangent cannot be drawn from any point outside the circle, but there can be only two tangents to a circle from a point outside the circle.
In the above figure, two tangents are drawn from the external point ‘R’. And note that the length of the tangents drawn from the external point to a circle is equal. Here AR=BR
Two parallel tangents at most for a given secant:
For every given secant of a circle, there are exactly two tangents which are parallel to it and touch the circle at two diametrically opposite points.
Theorems of Tangent to the Circle
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Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: In the figure beside XY is a tangent passing through point P.
Now, consider a point Q on XY and join OQ. And OP is the radius of the circle.
Proof: OQ>OP, this happens for every point on the line XY, except the point P. OP is the shortest of all distances from the point O to the points of XY.
Hence proved that OP is perpendicular to XY.
Theorem 2: The lengths of tangents drawn from an external point to a circle are equal.
Given: A circle with centre O. And P be an external point. PQ and PR are two tangents to the circle intersecting at points Q and R respectively.
Now join OQ, OR and OP.
Proof: As PQ is a tangent, OQ is perpendicular to OP (Tangents at any point of a circle is perpendicular to the radius through the point of contact)
So, ∠OQP=90°
Hence proved that Triangle OQP is the Right Triangle.
OR = OQ = Radius
And OP is common
So, Triangle OQP = Triangle ORP (RHS congruence)
So, we can say that the tangents PR and PQ are equal.
Example: Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ= 2 times the angle OPQ.
Solution: Given a circle with centre O and T is a point outside the circle. Tp and TQ are two tangents to the circle. P and Q are points of contact.
Sample Questions
Ques. 1: Prove that the lengths of tangents drawn from an external point to a circle are equal. (All India 2016, 4 Marks)
Ans: Given that- A circle C (O,r), P is a point outside the circle and PA and PB are tangents to a circle.
To Prove- PA=PB
Construction: Draw OA, OB and OP
Proof: Consider ∠OAPand ∠OBP
∠OAP = ∠OBP = 90° …(i)
OA = OB = Radius …(ii)
OP is common …(iii)
From eq. (i), (ii) and (iii)
We can say that,
∠OAP ≅ ∠OAP
Hence, It can be concluded that
PA = PB
Ques. 2: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (All India 2016, 4 Marks)
Ans: OA=OC (Radii of circle)
Now OB=OC+BC
∴OB>OC (OC being radius and B any point on tangent)
⇒OA<OB
B is an arbitrary point on the tangent.
Thus, OA is shorter than any other line segment joining O to any
point on tangent.
Since, the shortest distance of a point from a given line is the perpendicular distance from that line.
Hence, the tangent at any point of the circle is perpendicular to the radius.
Ques. 3: In figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS (All India- 2015, 4 Marks)
Ans: Given that, ∠RPQ=30o and PR and PQ are tangents drawn from P to the same circle.
Hence PR=PQ [Tangents drawn from an external point to a circle are equal in length]
∴ ∠PRQ=∠PQR [Angles opposite to equal sides are equal in a triangle. ]
In ∠PQR,
∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a triangle ]
⇒ 2∠RQP+30° =180°
⇒ 2∠RQP=150°
⇒ ∠RQP=75°
So,
∠RQP=∠QRP=75°
⇒ ∠RQP=∠RSQ=75° [ By Alternate Segment Theorem]
Given, RS || PQ
∴ ∠RQP=∠SRQ=75° [Alternate angles]
⇒ ∠RSQ=∠SRQ=75°
∴ QRS is also an isosceles triangle. [Sides opposite to equal angles of a triangle are equal.]
⇒ ∠RSQ + ∠SRQ + ∠RQS=180° [Angle sum property of a triangle]
⇒ 75° +75° +∠RQS=180°
⇒ 150° +∠RQS=180°
∴ ∠RQS=30°
Ques. 4: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (All India- 2015, 4 Marks)
Ans: OA=OC (Radii of circle)
Now OB=OC+BC
∴OB>OC (OC being radius and B any point on tangent)
⇒OA<OB
B is an arbitrary point on the tangent.
Thus, OA is shorter than any other line segment joining O to any
point on tangent.
Shortest distance of a point from a given line is the perpendicular distance from that line.
Hence, the tangent at any point of the circle is perpendicular to the radius.
Ques. 5: Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. (All India- 2014, 4 Marks)
Ans: To prove- AB || PT
Construction- Join OA,OB,&OP
Proof-
⇒OP⊥PT [Radius is ∠ to tangent through a point of contact]
⇒∠OPT=90°
Since P is the midpoint of Arc APB
⇒Arc(AAP)=arc(BP)
⇒∠AOP=∠BOP
⇒∠AOM=∠BOM
⇒In ΔAOM & ΔBOM
⇒OA=OB=r
⇒OM=OM(Common)
⇒∠AOM=∠BOM (proved above)
⇒∠AOM≅∠BOM (by SAS congruence axiom)
⇒∠AMO = ∠BMO
⇒∠AMO + ∠BMO=180°
⇒∠AMO = ∠BMO=90°
⇒∠BMO = ∠OPT=90°
But, they are corresponding angles.
Hence, we can say that AB || PT
Ques. 6: If given figure, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB. (All India- 2016, 2 Marks)
Ans: In ΔOAP and ΔOBP,
OP = OP (Common)
∠OAP = ∠OBP = 90° (Radius is perpendicular to the tangent at the point of contact)
OA = OB (Radius of the circle)
∴ ΔOAP ≅ ΔOBP (RHS congruence criterion)
⇒ ∠OAP = ∠OBP = 120°/2 = 60°
In ΔOAP,
cos ∠OPA = 60° = AP/OP
∴ 1/2 = AP/OP
⇒ OP = 2 AP
Ques. 7: In given figure, from a point P, two tangents PT and PS are drawn to a circle with centre O such that ∠SPT = 120°, Prove that OP = 2PS (All India- 2016, 2 Marks)
Ans: In ΔOTP and ΔOSP,
OP = OP (Common)
∠OTP = ∠OTP = 90° (Radius is perpendicular to the tangent at the point of contact)
OT = OS (Radius of the circle)
∴ ΔOTP ≅ ΔOSP (RHS congruence criterion)
⇒ ∠OTP = ∠OSP = 120°/2 = 60°
In ΔOTP,
cos ∠OPT = 60° = TP/OP
∴ 1/2 = TP/OP
⇒ OP = 2 TP
Ques. 8: In figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB. (All India- 2015, 1 Marks)
Ans: Given PA & PB are tangent to the circle with center O.
PA=PB [length of tangent from external point to circle are equal]
In ΔPAB
PA=PB
∠PBA=∠PAB [isosceles triangle]
now ∠PAB + ∠PBA + ∠APB= 180° [Angle sum property]
2∠PAB=180−50=130
∠PBA=∠PAB=65°………..(1)
Now PA is tangent & OA is radius at point A.
∠OAP=90°[tangent at any point is ⊥ to radius]
∠OAB=∠OAP−∠PAB=90−65=25°
Hence angle OAB is 25°.
Ques. 9: In figure, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ (All India- 2015, 2 Marks)
Ans: AB is the straight line
∠AOQ+∠BOQ=180°
∠BOQ=180°−58
∠BOQ=122°
In triangle BOQ,
OB=OQ
So ∠OBQ=∠OQB (Since sides opposite are equal angle opposite to the equal sides are equal)
So ∠OBQ + ∠OQB + ∠BOQ=180°
122o+2(∠OBQ)=180o→∠OBW=29°
In triangle ABT⇒∠ABT + ∠BAT + ∠BTA=180°= 29°+90°+∠BAT=180°
∠ATQ=61°
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