Theorem 10.2
prove that ” The Lengths of tangents drawn from an external points to a circle are equal.”
Given:
C ( O , r )
P = External point from the circle
P A and P B are tangents From External Point P
Construction:
Join External Point P , Point of contacts A and B to Center O .
Proof:
In Δ PAO and Δ PBO
P O = P O { Common Sides }
O A = O B { Radii Of Same Circle }
∠PAO = ∠PBO = 90º { Radius is always perpendicular (⊥) on Point of contact of tangent by Theorem 10.1 }
Δ PAO ≅ Δ PBO { By R.H.S. }
P A = P B { By CPCT }
Hence Proved
Key Points
-
- ∠OPA = ∠OPB { By CPCT }
- ∠POA = ∠POB { By CPCT }
- P O is the angle bisector of ∠APB and ∠AOB
Exercise – 10.2
In Question 1 to 3 , Choose the correct option and give justification.
Q. 1. From a point Q , the length of the tangent to a circle is 24cm and the distance of Q
From the center is 25cm.The radius of the circle is:
(A) 7cm
(B) 12cm
(C) 15cm
(D) 24.5cm
Answer – (C)
Explanation:
In Δ QPO
∠P = 90º
By PGT
(OP)² = (OQ)² – (PQ)²
=> (OP)² = (25)² – (24)²
=> (OP)² = 625 – 576
=> OP = √49
=> OP = 7
Hence Radius if circle = 7cm
Q.2. In figure 10.11, if T P and T Q are the two tangents to a circle with center O so that ∠POQ = 110º, Then ∠PTO is equal to :
(A) 60º
(B) 70º
(C) 80º
(D) 90º
Answer- (B)
According to figure
∠TPO = ∠TQO = 90º { By theorem 10.1 radius is perpendicular on point of contact to the tangent and circle }
∴ ∠POQ + ∠PTQ = 180º
=> 110º + ∠PTQ = 180º
=> ∠PTQ = 180º – 110º
=> ∠PTQ = 70º
Q.3. If tangents P A and P B From a Point P to a circle with center O are inclined to each other at angle of 80º, then ∠POA is equal to
(A) 50º
(B) 60º
(C) 70º
(D) 80º
Answer (A) 50º
Explanation:
According to figure
∠APB + ∠AOB = 180º
=> 80º + ∠AOB = 180º
=> ∠AOB = 180º – 80º
=> ∠AOB = 100º
=> ∠POA = ∠POB = 100º/2= 50º { ∠AOB/2 }
Q.4. Prove that the tangents drawn at the ends if a diameter of a circle are parallel.
Given:
According to figure
Circle ( O , r )
X Y and A B are tangents
To Prove:
X Y // A B { X Y parallel to A B }
Construction:
Join points of contacts P and Q to Center O.
Proof:
In figure
∠1 =∠2 { Radii is always perpendicular on point of contacts of tangent By theorem 10.1}
Here ∠1 = ∠2 Which is alternate interior angles and P Q is transversal on X Y and A B
∴ X Y // P Q
Hence the tangents drawn at the ends of a diameter of a circle are parallel.
Q.5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.
Given:
According to figure
C (O , r)
A B = tangent which is touches the circle at point P.
∴ P = Point of contact
To Prove:
The Perpendicular at the point of contact to the tangent to a circle passes through the center
Or
O P ⊥ on A B
Construction:
Join point of contact P to center O
Proof:
According to figure
OP = Radius of circle
A B = Tangent on circle
P = Point of contact on the circle
We know that by theorem 10.1
“Radius is always ⊥ ( perpendicular) on point of contact of tangent
∴ O P ⊥ on AB
Hence the perpendicular at the point of contact to the tangent to a circle passes through the center.
Q.6. The length of a tangent From a point A at distance 5cm From the center of the circle is 4cm. Find the radius of the circle.
Given:
According to figure
C ( O, r )
A B = Length of Tangent from point A = 4cm
O A = 5cm
To Find:
The radius of circle OB = r=?
Solution:
In ΔABO
∠ABO = 90º
A B = 4cm
A O = 5cm
O B =?
By PGT
(OA)² = (AB)² + (OB)²
=> (5)² = (4)² + (OB)²
=> 25 = 16 + (OB)²
=> 25 – 16 = (OB)²
=> 9 = (OB)²
=>√9 = OB
=> 3 = OB
Hence Radius of circle = 3cm
Q.7. Two concentric circles are of radii 5cm and 3cm.Find the length of the chord of the larger circle which touches the smaller circle.
Given:
According to Figure
Two concentric circles with radius O P = 3cm and O A = 5cm
To Find:
Chord A B =?
Solution:
According to condition
A B = Chord for Larger circle =Tangent for smaller circle
P = Point of contact
Now
In Δ APO
∠APO = 90º
A O = 5 cm
O P = 3cm
A P =?
By PGT
(AO)² = (AP)² + (3)²
=> (5)² = (AP)² + 9
=> 25 – 9 = (AP)²
=> 16 = (AP)²
=> √16 = AP
=> 4 = AP
Radius ( OP ) is always bisect the chord from the point of contact.
A P = P B = 4cm
∴ A B = AP + P B
=> A B = 4 + 4
=> A B = 8cm
Alternate Method:
In Δ APO and ΔBPO
A O = O B { Radii of same circle }
O P = O P {Common side }
∠APO = ∠BPO = 90º
=> Δ APO ≅ Δ BPO { RHS Congruency }
∴ A P = P B { By CPCT }
Hence A P = P B = 4cm
A B = A P + P B
=> A B = 4 + 4
=> A B = 8cm
∴ Length of the chord A B = 8cm.
Q.8. A Quadrilateral ABCD id drawn to circumscribe a circle. (see in figure 10.12)
Prove that A B + C D = A D + B C .
Given:
According to figure
A B , B C , C D and A D are tangents have point of contacts P , Q , R and S respectively.
To Prove:
A B + C D = A D + B C
Proof:
We know that length of tangent from an external point are equal.
∴ A P = A S ———Equation(i)
B P = B Q ———–Equation (ii)
C R = C Q ———–Equation(iii)
D R = D S ———–Equation(iv)
Adding Equation (i) + Equation (ii) + Equation(iii) +Equation(iv)
A P + B P + C R + D R = A S + B Q + C Q + D S
=> A B + CD = A S + D S + B Q + C Q
=> A B + C D = A D + B C
Hence Proved
Q.9. In Figure 10.13, X Y and X’ Y’ are two parallel tangents to a circle with center O and another tangent A B with point of contact C intersecting X Y at A and X’ Y’ at B. Prove that ∠AOB = 90º
Given:
According to figure
diagram
tangents X Y // X’ Y’
To Prove:
∠AOB = 90º
Construction:
Join O to C
Proof:
By Theorem 10.2 :We know that “Length of tangent Drawn From an External point are equal in Length”.
P A = A C ———– Equation (i)
In ΔQBO and ΔBCO
∠BQO = ∠BCO = 90º { By Theorem 10.1 Radius is always ⊥ on point of contact }
O B = O B {common side }
O Q = O C { Radii of same circle }
ΔQBO ≅ Δ BCO { By RHS Congruence }
∠1 = ∠2 ——————–(i)
Similarly
In Δ PAO and Δ CAO
P A = C A { From Equation (i) }
A O = A O { common side }
O P = O C { Radii of same circle }
∴ Δ PAO ≅ Δ CAO { By SSS congruence }
∠ PAO = ∠ CAO { By CPCPT }
=> ∠3 = ∠5 ———————(ii)
We know that sum of angles subtended same side on a line at a point are 180º
∠1 + ∠2 + ∠3 + ∠5 = 180°
=> ∠2 + ∠2 + ∠3 + ∠3 = 180º
=> 2∠2 + 2∠3 = 180º
=> 2(∠2 + ∠3) = 180º
=> ∠2 + ∠3 = 180º/2
=> ∠2 + ∠3 = 90º
=> ∠AOB = 90º
Hence Proved
Q.10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line – segment joining the points of contact at the center .
Given:
According to figure
di
P A and P B are tangents from an external point P
To Prove:
∠APB + ∠AOB = 180
Proof:
In figure
In Quadrilateral PAOB
∠PAO = ∠PBO = 90º { Radii is always ⊥ on point of contact }
∠APB + ∠PAO + ∠AOB + ∠PBO = 360º
=> ∠APB + 90º + ∠AOB + 90° = 360º
=> ∠APB + ∠AOB + 180º = 360º
=> ∠APB + ∠AOB = 360º – 180º
=> ∠APB + ∠AOB = 180º
Hence Proved
Q.11. Prove that the parallelogram circumscribing a circle is rhombus.
Given:
According to figure
Quadrilateral ABCD is a Parallelogram
A B // D C and A B =C D
A D // B C and A D = B C
Quadrilateral ABCD is circumscribing a circle with center O
To Prove:
Parallelogram ABCD is a rhombus
OR
A B = B C = C D = A D
Proof:
According to figure:
We know that By theorem 10.2 “Length of tangents drawn from an external point are equal”
B P = B Q ———–Equation(i)
A P = A S ————Equation (ii)
C R = C Q ———–Equation (iii)
D R = D S ———–Equation (iv)
Adding Equation (i) + Equation (ii) + Equation (iii) + Equation (iv)
B P + A P + C R + D R = B Q + A S + C Q + D S
=> A B + C D = B Q + C Q + A S + D S
=> A B + C D = B C + A D
=> 2 A B = 2 B C
=> A B = B C
∴ A B = B C = C D = A D
Hence parallelogram ABCD is a rhombus.
Q.12. A Triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segments B D and D C into which B C is divided by the point of contact D are of length 8cm and 6cm respectively (see fig. 10.14).Find the sides A B and A C.
Given:
In ΔABC:
ΔABC is drawn to circumscribe a circle if radius 4cm
To Find:
Length of side A B = ?
Length of side A C =?
Construction:
Join vertices A , B and C to center O
BC = 6 + 8 = 14 cm
We know that tangents drawn from an external point are equal in length.
∴ A F = A E = X (let)
C D = C F = 6cm
B D = B E = 8cm
a = A B = (x + 8 );
b = B C = 14 cm
c = A C = (6+x)
By Heron’s Formula
S = (A B + B C + A C)/2
=> S =(a + b + c)/2
=> S = ( x + 8 + 14 + 6 + x )/2
=> S = ( 2x + 28 )/2
=> S = 2(x + 14)/2
=> S = 2(x + 14)/2
=> S = x + 14
s – a = x + 14 -( x + 8)
=> s – a = x + 14 – x – 8
=> s – a = 6
s – b = x + 14 – 14
=> s – b = x
s – c = x + 14 – ( x + 6)
=> s – c = x + 14 – x – 6
=> s – c = 8
Area of Δ ABC = √s(s-a)(s-b)(s-c)
=> Area of Δ ABC = √(x+14)(x)(6)(8)
=> Area of Δ ABC = √(x+14)48x ————–Equation(i)
Area of Δ ABC = Area of Δ AOB + Area of Δ BOC + Area of Δ AOC
=> Area of Δ ABC = 1/2 × b × h + 1/2 × b × h + 1/2 × b × h
=> Area of Δ ABC = 1/2 × A B × 4 + 1/2 × B C × 4 + 1/2 × A C × 4
=> Area of Δ ABC = 1/2 ×4 ( A B + B C + A C )
=> Area of Δ ABC = 2 ( A B + B C + A C )
=> Area of Δ ABC = 2 ( 8 + x + 14 + x + 6 )
=> Area of Δ ABC = 2 (2x + 28)
=> Area of Δ ABC = 2×2(x + 14)
=> Area of Δ ABC = 4 ( x + 14) ———-Equation(ii)
Equation (i) = Equation (ii)
Area of Δ ABC = Area of Δ ABC
√(x+14)48x = 4 ( x + 14)
squaring both side
=> (x+14)48x = 16 ( x + 14)²
=> 48x = 16 ( x + 14)²/(x+14)
=> 48x = 16(x + 14)
=> 48x/16 = (x+ 14)
=> 3x = x + 14
=> 3x – x = 14
=> 2x = 14
=> x = 14/2
=> x = 7
Hence
A B = x + 8 = 7 + 8 = 15cm
B C = x + 6 = 6 + 7 = 13cm
Q.13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angle at the center of the circle.
Given:
According to Diagram:
ABCD is a quadrilateral circumscribing a circle with center O
tangents A B , B C , C D , and A D have point of contact P, Q, R and S respectively.
To Prove:
∠AOB + ∠COD = 180º
∠AOD + ∠BOC = 180º
Proof:
Angle subtend by the tangents external point on the center of circle are equal.
∠2 = ∠3 ——-Equation(i)
∠1 = ∠8 ——-Equation(ii)
∠4 = ∠5 ——-Equation(iii)
∠6 = ∠7 ——-Equation(iv)
We know that sum of all angles subtended on a point are 360°
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º
=> ∠1 + ∠2 + ∠2 + ∠5 + ∠5 + ∠6 + ∠6 + ∠1 = 360º { From Equation (i),(ii),(iii) and (iv) }
=> 2∠1 + 2∠2 + 2∠5 + 2∠6 = 360°
=> 2(∠1 + ∠2 + ∠5 + ∠6) = 360º
=> 2 ( ∠AOB + ∠COD ) = 360° { ∠1 + ∠2 = ∠AOB ; ∠5 + ∠6 = ∠COD }
=> ∠AOB + ∠COD = 360°/2
=> ∠AOB + ∠COD = 180º
Similarly
∠AOD + ∠BOC = 180º
Hence Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.