Theorem of Tangent to a Circle and Solved Examples - Some Applications of Trigonometry, Class 10 | Extra Documents, Videos & Tests for Class 10 PDF Download

PROPERTIES OF TANGENT TO A CIRCLE
Theorem-1 : The tangent at any point of a circle and the radius through the point are perpendicular to each other.
Given : A circle with centre O. AB is a tangent to the circle at a point P and OP is the radius through P.

To prove : OP  AB.
Construct : Take a point Q, other than P, on tangent AB. Join OQ.

Proof :

Hence, proved.

Remark 1: A pair of tangents drawn at two points of a circle are either parallel or they intersect each other at a point outside the circle.
Remark 2: If two tangents drawn to a circle are parallel to each other, then the line-segment joining their points of contact is a diameter of the circle.
Remark 3: The distance between two parallel tangents to a circle is equal to the diameter of the circle, i.e., twice the radius.
Remark 4: A pair of tangents drawn to a circle at the end points of a diameter of a circle are parallel to each other.
Remark 5: A pair of tangents drawn to a circle at the end points of a chord of the circle, other than a diameter, intersect each other at a point outside the circle.
Corollary 1: A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle
Given : O is the centre and r be the radius of the circle. OP is a radius of the circle. Line ℓ  is drawn through
P so that OP
To prove : Line  is tangent to the circle at P.

Construction : Suppose that the line  is not the tangent to the circle at P. Let us draw another straight line m which is tangent to the circle at P. Take two points A and B (other than P) on the line  and two points C and D on m.
Proof: 

But a part cannot, be equal to whole. This gives contradiction. Hence, our supposition is wrong.
Therefore, the line  is tangent to the circle at P
Corollary 2 : If O be the centre of a circle and tangents drawn to the circle at the points A and B of the circle intersect each other at P, then AOB + APB = 180°.

Proof:

Hence, Proved

Theorem-2 : If two tangents are drawn to a circle from an exterior point, then
(i) the tangents are equal in length
(ii) the tangents subtend equal angles at the centre
(iii) the tangents are equally inclined to the line joining the point and the centre of the circle.
Given : PA and PB are two tangents drawn to a circle with centre O, from an exterior point P.
To prove : (i) PA = PB (ii) AOP = BOP, (iii) APO = BPO.

Proof :

Corollary 3 : If PA and PB are two tangents from a point to a circle with centre 0 touching it at A and B Prove that OP is perpendicular bisector of AB.

Proof

Therefore, OP is perpendicular bisector of AB
Hence Proved.
COMMON TANGENTS OF TWO CIRCLES

Two circles in a plane, either intersect each other in two points or touch each other at a point or they neither intersect nor touch each other.
Common Tangent of two intersecting circles : Two circles intersect each other in two points A and B.
Here, PP' and QQ' are the only two common tangents. The case where the two circles are of unequal radii, we find the common tangents PP' and QQ' are not parallel.

Common tangents of two circles which touch each other externally at a point :

Two circles touch other externally at C.
Here, PP', QQ' and AB are the three common tangents drawn to the circles.
Common tangents of two circles which touch each other internally at a point :

Two circles touch other internally at C. Here, we have only one common tangent of the two circles.
Common tangents of two non-intersecting and non-touching circles :
Here, we observe that in figure (a), there is no common tangent but in figure (b) there are four common tangents PP', QQ', AA and BB'.

Ex.1 A point A is 26 cm away from the centre of a circle and the length of tangent drawn from A to the circle is 24 cm. Find the radius of the circle.
 Sol. Let O be the centre of the circle and let A be a point outside the circle such that OA = 26 cm.
Let AT be the tangent to the circle.

Then, AT = 24 cm. Join OT.
Since the radius through the point of contact is perpendicular
to the tangent, we have OTA = 90°. In right OTA, we have

 Ex.2 In the given figure, ΔABC is right-angled at B, in which AB = 15 cm and BC = 8 cm. A circle with centre O has been inscribed in ΔABC. Calculate the value of x, the radius of the inscribed circle.
 Sol. Let the inscribed circle touch the sides AB, BC and CA at P, Q and R respectively. Applying Pythagoras theorem
on right ΔABC, we have
AC² = AB² + BC² = (15)² + (8)² = (225 + 64) = 289

[OPB = 90°, PBQ = 90°, OQB = 90° and OP = OQ = x cm]
 BP = BQ = x cm.
Since the tangents to a circle from an exterior point are equal in length, we have AR = AP and CR = CQ.
Now, AR = AP = (AB – BP) = (15 – x) cm
CR = CQ = (BC – BQ) = (8 – x) cm.
 AC = AR + CR  17 = (15 – x) + (8 – x)  2x = 6  x = 3.
Hence, the radius of the inscribed circle is 3 cm.

Ex.3 If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
Sol. Let ABCD be a parallelogram whose sides AB, BC, CD and DA touch a circle at the points P, Q, R and S respectively.
Since the lengths of tangents drawn from an external point to a circle are equal, we have
AP = AS, BP = BQ, CR = CQ and DR = DS.

 AB + CD = AP + BP + CR + DR

= AS + BQ + CQ + DS
= (AS + DS) + (BQ + CQ)

 = AD + BC

Now, AB + CD = AD + BC

 2AB = 2BC [∵ Opposite sides of a ║ gm are equal]
 AB = BC
 AB = BC = CD = AD.

Hence, ABCD is a rhombus.

Ex.4 In the given figure, the incircle of ΔABC touches the sides AB, BC and CA at the points P, Q, R respectively.
Show that AP + BQ + CR = BP + CQ + AR = 1/2(Perimeter of ΔABC)

Sol. Since the lengths of two tangents drawn from an external point to a circle are equal, we have
AP = AR, BQ = BP and CR = CQ
 AP + BQ + CR = AR + BP + CQ ...(i)
Perimeter of ΔABC = AB + BC + CA = AP + BP + BQ + CQ + AR + CR

= (AP + BQ + CR) + (BP + CQ + AR)
= 2(AP + BQ + CR) [Using (i)]

 AP + BQ + CR = BP + CQ + AR =1/2(Perimeter of ΔABC)

Ex.5 In two concentric circles, prove that a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

Sol. Let there be two concentric circles, each with centre O.

Since OP is a radius of smaller circle and APB is a tangent to it at the point P, so OP ⊥  AB.
But the perpendicular from the centre to a chord, bisects the chord.
 AP = PB
Hence, AB is bisected at the point P.

Ex.6 Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Sol. Let O be the centre of the concentric circles and let AB be a chord of the outer circle, touching the inner circle at P. Join OA and OP.
Now, the radius through the point of contact is perpendicular to the tangent.
 OP  AB.
Since, the perpendicular from the centre to a chord, bisects the chord, AP = PB. Now, in right ΔOPA, we have
OA = 13 cm and OP = 5 cm.
 OP² + AP² = OA² AP² = OA² – OP² = (132 – 52) = (169 – 25) = 144.

Ex.7 In the given figure, PT is a common tangent to the circles touching externally at P and AB is another common tangent touching the circles at A and B. Prove that:
(i) T is the mid-point of AB
 (ii) APB = 90°
 (iii) If X and Y are centres of the two circles,

show that the circle on AB as diameter touches the line XY.

Sol. (i) Since the two tangents to a circle from an external point are equal, we have
TA = TP and TB = TP.
 TA = TB [Each equal to TP]

i)  

ii) Thus, P lies on the semi-circle with AB as diameter.
Hence, the circle on AD as diameter touches the line XY.

Ex.8 Two circles of radii 25cm and 9cm touch each other externally. Find the length of the direct common tangent.
Sol. Let the two circles with centres A and B and radii 25 cm and 9 cm respectively touch each other externally
at a point C.
Then, AB = AC + CB = (25 + 9) cm = 34 cm.
Let PQ be a direct common tangent to the two circles.
Join AP and BQ.
Then, AP  PQ and BQ  PQ.
[ Radius through point of contact is perpendicular to the tangent]

Draw, BL  AP.

Then, PLBQ is a rectangle.
Now, LP = BQ = 9 cm and PQ = BL.
 AL = (AP – LP) = (25 – 9) cm = 16 cm.
From right ΔALB, we have

AB² = AL² + BL² BL² = AB² – AL² = (34)² – (16)² = (34 + 16) (34 – 16) = 900

Ex.9 In the given figure, PQ = QR, RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of : (i) QOP (ii) QCP

Sol. (i) In ΔPQR,
PQ = QR PRQ = QPR    [s opp. to equal sides of a Δ are equal]
Also, QPR + RQP + PRQ = 180°  [Sum of the s of a Δ is 180°]
 68° + 2PRQ = 180°
 2PRQ = (180° – 68°) = 112°
PRQ = 56°.
QOP = 2PRQ = (2 × 56°) = 112°.   [Angle at the centre is double the angle on the circle]
(ii) Since the radius through the point of contact is perpendicular to the tangent, we have
OQC = 90° and OPC = 90°.

Now, OQC + QOP + OPC + QCP = 360° [Sum of the s of a quad. is 360°)
 90° + 112° + 90° + QCP = 360°.
QCP = (360° – 292°) = 68°.

Ex.10 With the vertices of ΔABC as centres, three circles are described, each touching the other two externally. If the sides of the triangle are 9 cm, 7 cm and 6 cm, find the radii of the circles.
 Sol. Let AB = 9 cm, BC = 7 cm and CA = 6 cm.
Let x, y, z be the radii of circles with centres A, B, C respectively.

Hence, the radii of circles with centres A, B, C are 4 cm, 5cm and 2 cm respectively.