Question

two circles with Centre O and p and radius 8 cm and 4 cm touch Each Other externally . find the length of their common tangent QR​

Answer 1

Answer:

Explanation: length of direct common tangent = (d^2 - (R1 - R2)^2)^1/2

where d = distance between two circle

 R1 = radius of first circle  ,

R2 = radius of second circle 

 length = (12^2 - 4^2)^1/2

          = 8√2 Cm

Answer 2

Given : two circles with Centre O and p and radius 8 cm and 4 cm touch Each Other externally .

To find : QR

Solution :

From the figure

Triangle OSP is a right angles triangle

and QR = PS

then

PQRS is a rectangle

since , The tangent is perpendicular to the radius  at the point of contact

and

PS is also perpendicular to OQ

thus

OS = OP -SP

OS = 8-4

OS = 4 cm

now , using PGT

$OP^2= OS^2+SP^2\\\\SP =\sqrt{OP^2-OS^2}\\\\SP =\sqrt{12^2-4^2}\\\\SP =\sqrt{128}\\\\SP=8\sqrt{2}cm$\

since , the opposite sides  of the rectangle are equal

therefore ,The length of the common tangent QR​ is 8√2 cm

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