Question

A point P is 18 cm from the centre of a circle. The radius of the circle is 12 cm. Find the length of the tangent drawn to the circle from the point P
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Answer 1

Answer: HOPE THIS WILL HELP YOU

Step-by-step explanation:

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Pythagoras Theorem has been used. We see that we are given the distance of the point P and radius of the Circle. Now if join all the points including Tangent, it forms a triangle. By theorem we know that Radius is perpendicular ⊥ to the Tangent where they meet. This can be seen from the diagram. Using this, we can find the length of Tangent.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}}$

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★ Solution :-

Given (including information from figure) ,

• Let the centre of the circle be O

• Let the point of contact of radius and tangent be point S

• Then, joining all the points POS forms a Triangle.

» Distance of Point P = OP = 18 cm

» Radius of circle = OS = 12 cm

» Length of the tangent = PS

✒ Theorem : When a tangent is produced from an external point to the circle and the radius is made to contact with tangent, then radius is perpendicular (⊥) to the tangent.

This means ∠OSP = 90° . Also, ∆POS is right angled triangle since ∟OSP is a right angle .

→ OS ⊥ PS

→ ∠OSP = 90°

✒ Pythagoras Theorem : The square of Hypotenuse is equal to the sum of squares of Base and Height.

From figure we see,

• Base = OS = 12 cm

• Hypotenuse = OP = 18 cm

• Height = PS

On applying Pythagoras Theorem here, we get

$\\\;\sf{:\rightarrow\;\;(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}$

By applying values, we get

$\\\;\sf{:\Longrightarrow\;\;(OP)^{2}\;=\;\bf{(OS)^{2}\;+\;(PS)^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;(18)^{2}\;=\;\bf{(12)^{2}\;+\;(PS)^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;324\;=\;\bf{144\;+\;(PS)^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;(PS)^{2}\;=\;\bf{324\;-\;144}}$

$\\\;\sf{:\Longrightarrow\;\;(PS)^{2}\;=\;\bf{180}}$

$\\\;\sf{:\Longrightarrow\;\;(PS)\;=\;\bf{\sqrt{180}}}$

$\\\;\sf{:\Longrightarrow\;\;(PS)\;=\;\bf{\sqrt{2\:\times\:2\:\times\:3\:\times\:3\:\times\:5}}}$

$\\\;\sf{:\Longrightarrow\;\;(PS)\;=\;\bf{(2\:\times\:3)\sqrt{5}}}$

$\\\;\bf{:\Longrightarrow\;\;(PS)\;=\;\bf{\orange{6\sqrt{5}\;\:cm}}}$

$\\\;\bf{:\Longrightarrow\;\;(PS)\;=\;\bf{\orange{13.42\;\;cm}}}$

*Note :: Here 6√5 ≈ 13.42 . So we can use any value for the answer.

$\\\;\underline{\boxed{\tt{Hence,\:\;length\;\:of\;\:tangent\;\:=\;\bf{\purple{13.42\;\:cm}\:\;\tt{or}\:\;\bf{\purple{6\sqrt{6}\;\:cm}}}}}}$

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★ More Formulas related to Circle ::

$\\\;\tt{\leadsto\;\;Area\;of\;Sector\;=\;\dfrac{\pi r^{2}\theta}{360^{\circ}}}$

$\\\;\tt{\leadsto\;\;Length\;of\;chord\;=\;\dfrac{2\pi r\theta}{360^{\circ}}}$

$\\\;\tt{\leadsto\;\;Area\;of\;Segment\;=\;\dfrac{\pi r^{2}\theta}{360^{\circ}}\;-\;\dfrac{1}{2}\:r^{2}\sin\theta}$

$\\\;\tt{\leadsto\;\;Area\;of\;Circle\;=\;\pi r^{2}}$

$\\\;\tt{\leadsto\;\;Perimeter\;of\;Circle\;=\;2\pi r}$

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