Question

In the figure shown, PAB is a common tangent to both the circles. C and D are the centres of the circles
If the radius of the two circles are 6r and 2r and CD = 12r, then find the length of AB and PA, in terms of r.
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Answer 1

Answer:

AB = 11.2r

PA = 5.6r

Step-by-step explanation:

ΔPBC is corresponding to the ΔPAD

Because of these three similarities property:

1) ∠PBC = ∠PAD = 90°

2) ∠CPB = ∠DPA (COMMON TO BOTH TRIANGLE)

3) ∠PCB = ∠PDA (If two angle are same then third will also be same)

CB = 6r

AD = 2r

CD = 12r

CE = 14r

CP = 14r + X

AB = Z

AP = Y

BP =X + Y

By corresponding triangle:

$\frac{AD}{DP} =\frac{BC}{CP}$

$\frac{2r}{2r + x} =\frac{6r}{14r + x}$

on solving x = 4r

Now    DP = 2r + 4r = 6r

        & CP = 18r

for ΔDPA

      $(DP)^{2} = (AD)^{2} + (AP)^{2}$

      $(6r)^{2} = (2r)^{2} + (Y)^{2}$

      $36r^{2} = 4r^{2} + Y^{2}$

      $Y^{2} = 32r^{2}$

      Y = 5.6r

by corresponding triangle

      $\frac{BC}{BP} =\frac{AD}{AP}$

      $\frac{6r}{5.6r + z} =\frac{2r}{5.6r}$

   z = 11.2r