Tangents $P A$ and $P B$ are drawn from an external point $P$ to two concentric circle with centre $O$ and radii $8 c m$ and $5 c m$ respectively, as shown in figure. If $A P = 15 c m,$ then find the length of $B P .$

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Solution

Given that: $O A = 8 c m, O B = 5 c m$ and $A P = 15 c m$
To find: $B P$
Construction: Join $O P .$
Therefore, tangent to a circle is perpendicular to the radius through the point of contact.
So, $O A ⊥ A P$ and $O B ⊥ B P$
On applying Pythagoras theorem in $Δ O A P$ , we obtain:
$(O P)^{2} = (O A)^{2} + (A P)^{2}$
$\Rightarrow (O P)^{2} = (8)^{2} + (15)^{2}$
$\Rightarrow (O P)^{2} = 64 + 225$
$\Rightarrow O P = \sqrt{289}$
$\Rightarrow O P = 17$
Thus, the length of $O P$ is $17 c m .$
On applying Pythagoras theorem in $Δ O B P$ , we obtain:
$(O P)^{2} = (O B)^{2} + (B P)^{2}$
$\Rightarrow (17)^{2} = (5)^{2} + (B P)^{2}$
$\Rightarrow 289 = 25 + (B P)^{2}$
$\Rightarrow B P = \sqrt{264}$
$\Rightarrow B P \approx 16.25$
Hence the length of $B P$ is $16.25 c m .$

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