Question

${\textbf{Subject : Mathematics}}$

${\textbf{Chapter : Tangency}}$


A point P is 16 cm from the centre of the circle. The length of tangent drawn from P to the circle is 12 cm. Find the radius of the circle.​

Answer 1

Hey !

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ɢɪᴠᴇɴ : ᴘ ɪs ᴀ ᴘᴏɪɴᴛ ᴀᴛ ᴅɪsᴛᴀɴᴄᴇ ᴏғ 16ᴄᴍ ғʀᴏᴍ ᴛʜᴇ ᴄᴇɴᴛʀᴇ ᴏғ ᴀ ᴄɪʀᴄʟᴇ ᴏ. ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏғ ᴛᴀɴɢᴇɴᴛ ᴅʀᴀᴡɴ ғʀᴏᴍ ᴘ ᴛᴏ ᴛʜᴇ ᴄɪʀᴄʟᴇ ɪs 12 ᴄᴍ.

ɪᴛ ɪs ʀᴇǫᴜɪʀᴇᴅ ᴛᴏ ᴍᴇᴀsᴜʀᴇ ᴛʜᴇ ʀᴀᴅɪᴜs ᴏғ ᴛʜᴇ ᴄɪʀᴄʟᴇ.

ɴᴏᴡ, ∠ᴏᴛᴘ= 90°

( ∵ ᴀ ᴛᴀɴɢᴇɴᴛ ᴛᴏ ᴀ ᴄɪʀᴄʟᴇ ɪs ⊥ ᴛᴏ ᴛʜᴇ ʀᴀᴅɪᴜs ᴛʜʀᴏᴜɢʜ ᴛʜᴇ ᴘᴏɪɴᴛ ᴏғ ᴄᴏɴᴛᴀᴄᴛ).

∴ ɪɴ ʀɪɢʜᴛ ᴀɴɢʟᴇ ᴛʀɪᴀɴɢʟᴇ ᴏᴛᴘ,

ᴏᴘ² = ᴏᴛ²+ ᴛᴘ²

ᴏʀ, 16² = ᴏᴛ² + (12)²

ᴏʀ, 256-144 = ᴏᴛ²

ᴏʀ, 112 = ᴏᴛ²

∴ ᴏᴛ = 10.89 ᴄᴍ

ʜᴇɴᴄᴇ, ʀᴀᴅɪᴜs = 10.89 ᴄᴍ

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Thanks !

Answer 2

Answer:

Given:

length of tangent = 12cm

distance from centre to the external point p = 16cm

we know that if we join the centre of circle to the point of contact of tangent

it forms a "right angles triangle"

let O be the centre of circle

p is external point

Q be the point of contact of tangent

$pqr \: is \: a \: right \: triangle \: right \: angled \: at \: q$

$\sqrt{ {16}^{2} - 12 ^{2} } = radius$

radius = 10.5

$\small\boxed{\fcolorbox{blue}{white}{THEREFORE\:THE\:RADIUS\:OF\:CIRCLE\:IS\:10.5cm}}$

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