Question

Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in Fig.3. If AP = 15 cm, then find the length of BP.

Answer 1

Given that : OA = 8 cm, OB = 5 cm and AP = 15 cm

To find : BP
Construction:Join OP.
Now, $OA\perp AP$ and $OB\perp BP$ $\because$ Tangent to a circle is perpendiuclar to the Radius through the point of contact
$\Rightarrow \angle OAP=\angle OBP={90}^{\circ }$
On applying Phthagoras theorem in $\Delta OAP$, we obtain:
$(OP{)}^2=(OA{)}^2+(AP{)}^2\Rightarrow (OP{)}^2=(8{)}^2+(15{)}^2\Rightarrow (OP{)}^2=64+225\Rightarrow OP=\sqrt{289}\Rightarrow (OP{)}^2=289\Rightarrow OP=17$
$\Rightarrow (OP{)}^2=289\Rightarrow OP=17$
Thus, the length of OP IS 17 cm
On appling Phthagoras theorem in $\Delta OBP$, we obtain :
$(OP{)}^2=(OB{)}^2+(BP{)}^2$
$\Rightarrow (17{)}^2=(5{)}^2+(BP{)}^2\Rightarrow 289=25+(BP{)}^2\Rightarrow (BP{)}^2=289-25\Rightarrow (BP{)}^2=264$
$\Rightarrow BP=16.25cm$ (approx)