Question

In the given figure, AB and BC are the tangents to the circle from the point B. D is the centre of the circle. BD = 5 cm and
CD = 3 cm. Find the value of AB - BC.

• A. 20 cm
• B. 4 cm
• C. 9 cm
• D. 0 cm

Answer 1

The correct option is D

0 cm

In the diagram given in question,
AD = DC = 3 cm (Radius of the circle)

In $\Delta ADB$
AD = 3 cm (Radius of the given circle)
DB = 5 cm (Given)

By Theorem -The tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\therefore$ $\angle$DAB = ${90}^{\circ }$

Now, applying pythagoras theorem in right angled $△$DAB
$(AB{)}^2+(AD{)}^2=\left(BD{)}^2\right(AB{)}^2+3^2=5^2(AB{)}^2=25-9(AB{)}^2=16$
$AB$ = 4

$AB$ = $BC$ = 4 ( By Theorem - If two tangents are drawn to a circle from an exterior point, the tangents are equal in length)

$\therefore$ $AB$ - $BC$ = 4 - 4 = 0 cm