Question

If the radius of two concentric circles are r1 & r2 , where r2 is greater than r1 and a is the
length of the chord of the bigger circle which is tangent to the smaller circle.
Show that 4r22 = a2 + 4r12
​

Answer 1

Step-by-step explanation:

Given:- two concentric circles with centre O, whose radius were r1 and r2 and r2 > r1 and has larger circle has a chord AB = a and AB is tangent to the smaller circle.

To prove:- 4(r2)² = a² + 4(r1)²

Construction:- Draw OC perpendicular to AB and draw O to A which forms OA

Proof:-

Let r1 = r and r2 = R

we know that chord AB is tangent to the smaller circle joining O to C

so OC = r

also OA = R

we also know that the radius is perpendicular to the tangent (Theorem)

so triangle OAC is a right triangle

also, AB is a chord of the larger circle so a line drawn perpendicular from the centre bisects the chord.(Theorem)

so AC = a/2

Now in Triangle OAC,

By pythagoras theorem

r² + (a/2)² = R²

r² + (a²/4) = R²

Now let's multiply by 4 to avoid fractions,

4r² + 4(a²/4) = 4R²

4r² + a² = 4R²

Now we said that r = r1 and R = r2

so,

4(r1)² + a² = 4(r2)²

Hence proved

Hope you understood it........All the best