Write 'True' or 'False' and give reasons for your answer.
If the angle between two tangents drawn from a point $P$ to a circle of radius $a$ and center $O$ is $90^{\circ}$,then $OP = a\sqrt{2}$.
If the angle between two tangents drawn from a point $P$ to a circle of radius $a$ and center $O$ is $90^{\circ}$,then $OP = a\sqrt{2}$.
(A) True
Let the two tangents from point $P$ touch the circle at points $T$ and $R$. Given,the radius $OT = a$.
The line segment $OP$ bisects the angle between the two tangents,$\angle T P R$.
Therefore,$\angle T P O = \angle R P O = \frac{90^{\circ}}{2} = 45^{\circ}$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OT \perp PT$.
In the right-angled triangle $\triangle OTP$,we have:
$\sin 45^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OT}{OP}$
Substituting the values,we get:
$\frac{1}{\sqrt{2}} = \frac{a}{OP}$
Therefore,$OP = a\sqrt{2}$.
Let the two tangents from point $P$ touch the circle at points $T$ and $R$. Given,the radius $OT = a$.
The line segment $OP$ bisects the angle between the two tangents,$\angle T P R$.
Therefore,$\angle T P O = \angle R P O = \frac{90^{\circ}}{2} = 45^{\circ}$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OT \perp PT$.
In the right-angled triangle $\triangle OTP$,we have:
$\sin 45^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OT}{OP}$
Substituting the values,we get:
$\frac{1}{\sqrt{2}} = \frac{a}{OP}$
Therefore,$OP = a\sqrt{2}$.
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