In Fig. 9.19, the common tangent, AB and CD to two circles with centres O and O' intersect at E. Prove that the points O, E, O' are collinear
Solution:
Given, the tangents AB and CD are common to two circles with centres O and O’
The common tangents AB and CD intersect at E.
We have to prove that the points O, E and O’ are collinear.
Join OA and OC, O’D and O’B
Considering triangles AOE and EOC,
OE = OE = common side
OA = OC = radius of circle
We know that the radius of a circle is perpendicular to the tangent at the point of contact.
Tangents from external point E are EA and EC
So, AE = EC
By SAS criterion, the triangles AOE and EOC are similar.
Corresponding parts of congruent triangles or cpct is used to denote the relation between the sides and the angles of two congruent triangles.
By cpct,
∠AEO = ∠CEO ------------------ (1)
From the figure,
∠AEC = ∠AEO + ∠CEO
From (1), ∠AEC = ∠AEO + ∠AEO
∠AEC = 2∠AEO ------------------- (2)
From the figure,
CD is a straight line
The linear pair of angles ∠AED + ∠AEC = 180°
From (2), 2∠AEO + ∠AED = 180°
Dividing by 2,
∠AEO + (1/2)∠AED = 90°
∠AEO = 90° - (1/2)∠AED --------------- (3)
Considering triangles O’ED and O’EB,
O’B = O’D = radius of circle
O’E = O’E = common side
We know that the tangents drawn to a circle through an external point are equal.
So, EB = ED
By SSS criterion, the triangles O’ED and O’EB are similar.
By cpct, ∠O’ED = ∠O’EB ------------ (4)
From the figure,
∠DEB = ∠O’EB + ∠O’ED
From (4), ∠DEB = ∠O’ED + ∠O’ED
∠DEB = 2∠O’ED ------------- (5)
From the figure,
AB is a straight line
The linear pair of angles ∠AED + ∠DEB = 180°
From (5), ∠AED + 2∠O’ED = 180°
Dividing by 2,
(1/2)∠AED + ∠O’ED = 90°
∠O’ED = 90° - (1/2)∠AED -------------- (6)
Adding ∠AEO + ∠AED + ∠O’ED ,
From (3) and (6),
= 90° - (1/2)∠AED + ∠AED + 90° - (1/2)∠AED
= 180° - ∠AED + ∠AED
= 180°
So, ∠AEO + ∠AED + ∠O’ED = 180°
The points O, E and O’ lie on the same line.
Therefore, the points E, O and O’ are collinear.
✦ Try This: Two circles with centers O and O’ intersect at two points A and B. A line PQ is drawn parallel to OO’ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO’.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.4 Problem 10
In Fig. 9.19, the common tangent, AB and CD to two circles with centres O and O' intersect at E. Prove that the points O, E, O' are collinear
Summary:
In Fig. 9.19, the common tangent, AB and CD to two circles with centres O and O' intersect at E. It is proven that the points O, E, O' are collinear
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