Question

Draw a circle with centre O and radius 6cm. take a point P outside the circle at a distance of 10cm from O. Draw tangents to the circle from the point P. let the tangents intersect the circle in points A and B, Find BP.

A) 6cm
B) 8cm
C) 8.5cm
D) None of these

Answer 1

Hint: To find the value of BP, we’ll first take the help of the property of tangent that it is always perpendicular to the radius of the circle at the point of tangency.

Now, we’ll apply the Pythagoras theorem on the triangle that has BP as it’s one of the sides, using this we’ll get the value of BP as required.

Complete step by step solution: Given data: $radius(r) = 6cm$ and $PO = 10cm$

We know that tangents made from any outside point to a circle are always perpendicular to the radius of the circle at the point of tangency.

Therefore we can say that $OB \bot PB$

It is well known that if in triangle LMN $\angle L = {90^ \circ }$ then $M{N^2} = L{M^2} + L{N^2}$

And this is known as Pythagoras theorem

Now, using Pythagoras theorem in triangle PBO as $\angle B = {90^ \circ }$

$ \Rightarrow P{O^2} = O{B^2} + P{B^2}$

Now substituting the value of PO and radius as OB

$ \Rightarrow {10^2} = {6^2} + P{B^2}$

$ \Rightarrow P{B^2} = {10^2} - {6^2}$

Solving the right-hand expression

$ \Rightarrow P{B^2} = 100 - 36$

$ \Rightarrow P{B^2} = 64$

Taking square root on both sides

$\therefore PB = 8cm$

Option(B) is correct.

Note: Here we have given the tangent of the circle, so let us discuss some properties related to the tangent of a circle.

1) A tangent of a circle always touches the circle at a single point.
2) Tangent is always perpendicular to the radius made at the point of tangency.
3) The length of two tangents drawn to a single point to a circle is always equal.