Question

Two circles of radii 8 cm and 3 cm have their centers 13 cm apart. Find the length of a direct common tangent to the two circles.

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Answer 1

Given that AP = 8 cm, BQ = 3 cm and PQ = 13 cm

AB is the common tangent to the two circles.

We know that PA and QB are $\perp$ AB

Draw QR $\perp$ AP

ARQB forms a rectangle

AR = BQ = 3 cm

PR = AP – AR

PR = 8 – 3 = 5 cm

In $△$PQR, applying Pythagoras theorem we get

${QR}^2$ = ${PQ}^2$ – ${RP}^2$

${QR}^2$ = ${13}^2$ – $5^2$

${QR}^2$ = 169 – 25

${QR}^2$ = 144

QR = 12

AB = QR = 12 cm i.e. the length of common tangent.