If d₁, d₂ (d₂ > d₁) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, prove that d₂² = c² + d₁²

Solution:

Given, d₁ and d₂ are the diameters of two concentric circles.

c is the length of the chord of a circle which is tangent to the other circle.

We have to prove that d₂² = c² + d₁²

From the figure,

Let O be the centre of two concentric circles.

Let AB be the chord of the outer circle whose length is M

The chord AB is the tangent to the inner circle at the point D

Radius of circles = d₁/2 and d₂/2

Considering triangle OAB,

OA = OB = radius of the outer circle

Since the two sides are equal, OAB is an isosceles triangle.

We know that the radius of the circle is perpendicular to the tangent at the point of contact.

So, OD ⟂ AB

We know that the perpendicular from the centre of a circle to the chord always bisects the chord.

So, AD = DB = c/2

Considering triangle ODB,

∠ODB = 90° as the radius is perpendicular to the tangent at the point of contact.

So, ODB is a right triangle with D at right angle.

By using pythagoras theorem,

OB² = OD² + BD²

OD = radius of inner circle

OB = radius of outer circle

(d₂/2)² = (d₁/2) + (c/2)²

d₂²/4 = d₁²/4 + c²/4

Cancelling out common term,

d₂² = d₁² + c²

Therefore, it is proved that d₂² = d₁² + c²

✦ Try This: If d₁, d₂, d₃ are the diameters of the three escribed circles of a triangle, then d₁d₂ + d₂d₃ + d₃d₁ is equal to

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10

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NCERT Exemplar Class 10 Maths Exercise 9.3 Sample Problem 1

If d₁, d₂ (d₂ > d₁) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, prove that d₂² = c² + d₁²

Summary:

If d₁, d₂ (d₂ > d₁) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle. It is proven that d₂² = c² + d₁²

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