Question

If the radius of the circle is 5 cm and distance from centre to the point of intersection of two tangents is 13 cm. Find length of the tangent.
A.11 cm
B.10 cm
C.12 cm
D.13 cm

Answer 1

Hint: Here we consider the property i.e..The radius of a circle is perpendicular to its tangent and Pythagoras theorem: Let us consider a triangle \[ABC\] whose \[\angle B\] is right angle. Then, it states that, \[Hypotenuse{e^2} = Bas{e^2} + Perpendicula{r^2}\]

Complete step by step answer:

                           

Let us consider, O be the center of the circle. And AM and AN be two tangents which meets at A.
Now according to the problem, \[OA = 13\]and the radius is \[5\]cm.
So, \[OM = ON = 5\]
We have to find the length of the tangent that is to find the length of \[AM.\]
Let us consider the triangle \[AOM\].
We know that the radius of a circle is perpendicular to its tangent.
Then, \[OM \bot OA\]. So, triangle \[AOM\]is a right-angle triangle whose, \[\angle OMA = {90^ \circ }\].
We can say that, \[OM\]is the perpendicular, \[AM\] is the base and \[AO\] is the hypotenuse.
Then by Pythagoras theorem we can further say that,
\[A{O^2} = O{M^2} + A{M^2}\]
Now let us substitute the values of \[OA = 13\] and \[OM = 5\] to find AM,
Therefore, we get,
\[{13^2} = {5^2} + A{M^2}\]
Now we are going to solve the above equation to get AM,
\[A{M^2} = 169 - 25\]
We should solve again to find AM,
\[AM = \sqrt {144} = 12\]
Hence, the length of the tangent is found to be \[12\]cm.
The correct option is (C)\[12\] cm.

Note-: Here, \[AM = \pm \sqrt {144} = \pm 12\]
We take only the positive value since the length cannot be negative. Here we can also use the right angle triangle AON to find the length of the tangent.