Question

Two circles with centres O and P, and radii 8 cm and 4 cm touch each other externally. Find the length of their common tangent QR.

• A. $8$ cm
• B. $7$ cm
• C. $8\sqrt{2}$ cm
• D. $7\sqrt{3}$ cm

Answer 1

The correct option is C $8\sqrt{2}$ cm

The common tangent $QR$ is equal to $PS$

Triangle OSP is right angled triangle. (from given figure)

PSQR is a rectangle. (All four interior angles of quadrilateral PSQR are right angles. Since tangent is perpendicular to radius at point of contact; and PS is also perpendicular to OQ)

$OS=OP-SP=8-4=4cm$

From the Pythagorean Theorem we have

${OP}^2={OS}^2+{SP}^2$

now $SP=\sqrt{{OP}^2-{OS}^2}$

$SP=\sqrt{{12}^2-4^2}$...... [$OS=4cm$ and $OP=8+4=12cm$]

$SP=\sqrt{128}$

$SP=8\sqrt{2}cm$

also the opposite sides of a rectangle are equal .

Therefore length of commen tangent $QR=8\sqrt{2}cm$

So, option C is the answer.