In Fig. 9.13, AB and CD are common tangents to two circles. If the radii of the two circles are equal, prove that AB = CD
Solution:
Given, AB and CD are the common tangents of two circles with equal radii.
We have to prove that AB = CD
We know that the radius of the circle is perpendicular to the tangent at the point of contact.
So, OA ⟂ AB, O'D ⟂ CD, O'B ⟂ AB and OC ⟂ CD
Also, ∠OAB = ∠O'BA = ∠OCD = ∠ODC = 90°
From the figure,
We observe that AC and BD are straight lines
ABCD represents a quadrilateral
∠A = ∠B = ∠C= ∠D = 90°
So, ABCD is a rectangle.
We know that the opposite sides are equal in a rectangle.
i.e., AB = CD and AC = BD
Therefore, it is proved that AB = CD
✦ Try This: AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to
Given, AB and CD are two common tangents to circles which touch each other at C.
D lies on AB such that CD = 4 cm
We have to find the length of AB.
We know that the tangents drawn through an external point to a circle are equal.
From the figure,
D is the external point
DA, DB and DC are the tangents drawn to the circle which touch each other at C.
So, DA = DB = DC = 4 cm
We know that AB = AD + DB
So, AB = 4 + 4
AB = 8 cm
Therefore, the length of AB is 8 cm.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 6
In Fig. 9.13, AB and CD are common tangents to two circles. If the radii of the two circles are equal, prove that AB = CD
Summary:
In Fig. 9.13, AB and CD are common tangents to two circles. If the radii of the two circles are equal. It is proven that AB = CD
☛ Related Questions:
- If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠D . . . .
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- In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD