Question

the length of the tangants from a point A at a circle of radius 3 cm is 4 cm . the distance of A from the centre of the circle is​

Answer 1

Given:

• The length of the tangants from a point A at a circle of radius 3 cm is 4 cm.

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To find:

• The distance of A from the centre of the circle is?

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Solution:

• Let O be the center of circle.
• Let A be the external point from which tangent is drawn.
• Let R be the point of intersection of radius and tangent

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$\setlength{\unitlength}{1.1mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\line(1,3){6.4}}\put(20,52.8){\line(3, - 1){68.5}}\put(25,30){\circle*{1}}\put(88,30){\circle*{1.2}}\put(22,28){\sf O}\put(18,39){\sf 3 cm}\put(31,52){\sf R}\put(90,28.5){\sf A}\put(25,30){\line(2,0){63}}\put(33,45){\line(1,3){1.1}}\put(30,45.8){\line(3, - 1){3.2}}\put(60,42){\sf 4 cm}\end{picture}$

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∆ ARO is a right angled triangle,

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where,

• OR = 3 cm
• AR = 4 cm

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Using Pythagoras Theorem,

⇏ AO² = AR² + OR²

⇏ AO² = 4² + 3²

⇏ AO² = 16 + 9

⇏ AO² = 25

⇏ AO = 5 cm

Hence, the distance of A from the centre of the circle is 5 cm.

Answer 2

Solution :-

Given,

• the length of the tangent from a point a at a circle of radius 3cm and 4cm .

To find ,

• the distance of a from the centre of the circle ?

Now,

• let O be the centre .
• and a the external point .
• and T point of intersection .

according to the figure it form a right angle triangle

So, angel OAT = 90°

by using the Pythagoras theorem ;

$\boxed{{ac}^{2} = {ab}^{2} + {bc}^{2} }$

putting the value we get ;

$= > {ao}^{2} = {ab}^{2} + {or}^{2}$

$= > {ao}^{2} = {4}^{2} + {3}^{2}$

$= > {ao}^{2} = 16 + 9$

$= > {ao}^{2} = 25$

$= > ao = \sqrt{25}$

$= > ao = 5cm$

The distance of a from the centre of circle is 5 cm .