Question

In the figure, B and C are the centres of the 2 circles with radii 9 cm and 3 cm respectively. Also, PQ is the transverse common tangent.
Find the length of PQ, if the centres of circles are 15 cm apart.

• A. 9 cm
• B. 2 cm
• C. 10 cm
• D. 8 cm

Answer 1

The correct option is A

9 cm

Draw MC $\perp$ MB

$PQ\perp PB$

$CQ\perp PQ$

$\angle QPM=\angle PMQ=\angle CQP={90}^{\circ }$

Since PMCQ is a quadrilateral, sum of angles $={360}^{\circ }$

$\therefore \angle MCQ={90}^{\circ }\therefore \angle MPQ=\angle PQC=\angle QCM=\angle PMC={90}^{\circ }$

$\therefore$ PMCQ is a rectangle.

PM = QC = 3 cm

PQ = MC

Now, BM = BP + PM = 9 + 3 = 12 cm

In the right angled triangle BMC,

$B{C}^2=B{M}^2+M{C}^2$

Given, BC = 15

${15}^2={12}^2+M{C}^{2}225=144+M{C}^{2}225-144=(MC{)}^{2}81=(MC{)}^2\sqrt{81}=MC$

$\therefore$ MC = 9
MC = PQ = 9
$\therefore$ Legth of transverse common tangent = 9 cm.