Question

12. In the adjoining figure, if AP = 8 and BP = 6,find the ratio of the radii of the two circles?(1) 16:9(2) 4:3(3) 25:9(4) 25 : 16​

Answer 1

Given : AP = 8 & BP = 6

To find : ratio of the radii of the two circles

Solution:

AQ = QP    ( equal Tangent )

BQ = QP  ( equal tangent )

AP = QP = BQ =  x

in ΔAPQ     AP² = AQ² + QP²  - 2AQ.QPCosα  

=> x² + x²  - 2x²Cosα = 8²

α = ∠AQP    then 180 - α  = ∠BQP

in ΔBPQ     BP² = BQ² + QP²  - 2BQ.QPCos(180 -α  )

=> x² + x²  - 2x²Cosα = 6²

=> 4x² = 100

x = 5

QP  = 5

Join Q with center of circle intersecting AP at M   &  PB at N

QM = √(QP² -  (AP/2)²) = √5² - (8/2)²   = 3

QN =  √(QP² -  (BP/2)²) = √5² - (6/2)²   = 4

Join P with center  

QP ² + PO²  =  (QM + OM)²    PO = Radius = R

OM  = √PO² - (PM)²  = √PO² - (AP/2)²  = √(R² - 4²)

=> 5² + R²  = ( 3 + √(R² - 4²) ) ²  

=>  25 + R² = 9  + R² - 16  + 6√(R² -16)

=> 32/6 = √(R² -16)

=> 16/3  = √(R² -16)

=> 256/9 = R² -16

=> 400/9 = R²

=>  R = 20/3

Similarly in other circle

Join P with center  O'

QP ² + PO'²  =  (QN + O'N)²    PO' = Radius = r

O'N  = √PO'² - (PN)²  = √PO'² - (BP/2)²  = √(r² - 3²)

=> 5² + r²  = ( 4 + √(r² - 3²) ) ²  

=>  25 + r² = 16  + r² - 9  + 8√(r² -9)

=> 18/8 = √(r² -9)

=> 9/4  = √(r² -9)

=> 81/16 = r² -9

=> 225/16 = r²

=>  r = 15/4

R : r     20/3 : 15/4

=> 80 : 45

=> 16 : 9

option 1 is correct

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