Question

The length a tangent from a point A at a distance 5 cm from the centre of a circle is 4

cm, find the radius of the circle. ​

Answer 1

${ \huge{ \red{ \underline{ \bf{Given : }}}}}$

• AB is the tangent from point A

• Length of tangent = AB = 4 cm

• Also, Distance of point from circle = 5 cm. Hence, OA = 5 cm.

${ \huge{ \green{ \underline{ \bf{To \: Find :}}}}}$

• Radius of the circle ?

${ \huge{ \underline { \bf{ \blue{Using \: Pythagoras \: Theorem :}}}}}$

• (Hypotenuse)² = ( Height)² + (Base)².

${ \huge{ \underline{ \red{ \bf{Solution :}}}}}$

$↬{ \mathsf{Let \: the \: circle \: be \: with \: center \: O}}$

$↬{ \mathsf{Since \: AB \: is \: tangent}}$

$↬{ \mathsf{Hence \: OB \: ⊥ \: AB}}$

$↬∴ { \mathsf{\angle OBA = 90 \degree}}$

$↬{ \mathsf{So, \: \triangle OAB \: is \: a \: right \: triangle}}$

${ \huge{ \purple{ \underline{ \bf{Then :}}}}}$

$</p><p></p><p>✠ { \mathsf{(Hypotenuse)² \: = \: ( Height)² \: + \: (Base)²}}</p><p>$

$⇒{ \mathsf{OA {}^{2} = OB {}^{2} + AB {}^{2} }}$

$⇒{ \mathsf{5 {}^{2} = OB {}^{2} + 4 {}^{2}}}$

$⇒{ \mathsf{OB {}^{2} = 25 - 16}}$

$⇒{ \mathsf{ OB {}^{2} = 9}}$

$⇒{ \mathsf{OB = \sqrt{9} }}$

$⇒{ \mathsf{OB = \sqrt{3 {}^{2} }}}$

$⇒{ \mathsf{OB = 3}}$

${ \huge { \pink { \underline{ \bf{Therefore,}}}}}$

${ \mathsf{ \: Radius \: of \: the \: circle = { \boxed{ \bf{ \orange{OB = 3 cm}}}}}} \star$

Answer 2

$\large\mathfrak\colorbox{plum}{Answer}$

Given,

• AB is the tangent from point A
• Length of tangent = AB = 4 cm
• Also, Distance of point from circle = 5 cm. Hence, OA = 5 cm.

To find,

Radius of Circle

Solution,

Let the circle be with center O

Since,AB is a tangent

Hence, OB ⊥ AB

∴∠OBA=90°

So,△OAB is a right angled triangle.

Applying Pythagores Theorem,

$\normalsize\orange{\boxed{\sf{(Hypoteneus)²=(Height)²+(Base)²}}}$

$\normalsize\sf{OA²=OB²+AB²}$

$\normalsize\sf{5²=OB²+4²}$

$\normalsize\sf{25=OB²+16}$

$\normalsize\sf{OB²=25-16}$

$\normalsize\sf{OB²=9}$

$\normalsize\sf OB = \sqrt{9}$

$\large\pink{\boxed{\sf{Radius\: of\: circle=3}}}$