Question

The distance between the centres of equal circles each of radius 3 cm is 10 cm. The length of a transverse tangent AB is

• A. 4 cm
• B. 6 cm
• C. 8 cm
• D. 10 cm

Answer 1

The correct option is C

8 cm

∠OAC = ∠CBP = ${90}^{\circ }$
(Tangent Perpendicular to Radius at Point of Contact)

∠OCA = ∠PCB (Vertically opposite angle)

Triangle OAC is similar to PBC

$\frac{OA}{PB}$ = $\frac{OC}{PC}$

$\frac{3}{3}$ = $\frac{OC}{PC}$

OC = PC

But PO = 10 cm

Therefore OC = PC = 5cm

${AC}^2$ = ${OC}^2$ – ${OA}^2$

${AC}^2$ = $5^2$ – $3^2$

AC = 4 cm

Similarly BC = 4 cm

$\therefore$ AB = 8 cm