Question

In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.

Answer 1

i. PQ = RQ

∴ ∠PRQ = ∠QPR  ...(Opposite angles of equal sides of a triangle)

`=>` ∠PRQ + ∠QPR + 68° = 180°

`=>` 2∠PRQ = 180° – 68°

`=> ∠PRQ = (112^circ)/2 = 56^circ`

Now, ∠QOP = 2∠PRQ  ...(Angle at the centre is double)

`=>` QOP = 2 × 56° = 112°

ii. ∠PQC = ∠PRQ  ...(Angles in alternate segments are equal)

∠QPC = ∠PRQ  ...(Angles in alternate segments)

∴ ∠PQC = ∠QPC = 56°  ...(∵ ∠PRQ = 56° from(i))

∠PQC + ∠QPC + ∠QCP = 180°

`=>` 56° + 56° + ∠QCP = 180°

`=>` ∠QCP = 68°

Answer 2

Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.

To prove the given question, it is sufficient to prove AB = CD.

For this join OM, ON, OB and OD.

Let the radius of outer and inner circles be R and r respectively.

AB touches the inner circle at M.

 AB is a tangent to the inner circle 

∴ OM ⊥ AB

`=>` BM = `1/2` AB

`=>` AB = 2BM

Similarly ON ⊥ CD, and CD = 2DN

Using pythagoras theorem in ΔOMB and ΔOND

OB² = OM² + BM², OD² = ON² + DM²

`=> "BM" = sqrt ("R"^2 - "r"^2), "DN" = sqrt ("R"^2 - "r"^2)`

Now, 

AB = 2BM = `2 sqrt ("R"^2 - "r"^2) , "CD" = 2"DN" = 2 sqrt ("R"^2 - "r"^2)` 

∴ AB = CD

Hence Proved.