Question

A tangent PQ at a point P of a circle of radius $5cm$ meets a line through the center O at a point Q so that $OQ = 12cm$ the Length of a PQ is
A.\[12\] cm
B.$13$ cm
C.$8.5$ cm
D.$\sqrt {119} $ cm

Answer 1

Hint: Here We are asked to find the length of a PQ. First, draw a figure by using the given conditions radius $r = 5cm$ which meets the line through the center O at a point Q so that $OQ = 12cm$ and solve this question by using Pythagoras Theorem.

Complete answer:
Let us first draw a diagram using the given data for a clear look.

The line drawn from the center O of the circle to the tangent is perpendicular to the tangent
that is $OP \bot PQ$
Here, $OP = 5cm$ which is the radius of the circle, and $OQ = 12cm$ .
From the figure, $\angle OPQ = {90^ \circ }$ (Angle between a tangent and radius through the point of contact is ${90^ \circ }$)
Hence, $\Delta OPQ$ forms a right-angled triangle.
Now let us find the find the length of $PQ$
Let us consider $\Delta OPQ$ , right angled at $P$
Using Pythagoras Theorem, we can say that,
${\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {height} \right)^2}$
$O{Q^2} = P{Q^2} + O{P^2} $
$\therefore P{Q^2} = O{Q^2} - O{P^2}$
$PQ = \sqrt {O{Q^2} - O{P^2}}$
$= \sqrt {{{12}^2} - {5^2}}$
$= \sqrt {144 - 25} $
$= \sqrt {119} cm$
Therefore, $PQ$ = $\sqrt {119} cm$
Hence the length of $PQ$ is $\sqrt {119} cm$

The correct answer is option (D).

Note:
In this problem, one of the most important things that we need to know is Pythagorean theorem. Since from the given data we have formed a right-angle triangle since we know the length of two sides of that triangle so it is easy to find the third side by using the Pythagorean theorem. Student must give more attention while converting the given data into a diagram if the diagram goes wrong then the whole problem goes wrong.