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:-

Let the R be the radius and O be the centre of the circle on which PT is a tangent from point P

In Triangle OTP,

${ot}^{2} + {pt}^{2} = {op}^{2} \\ {r}^{2} + {12}^{2} = {16}^{2} \\ {r}^{2} = 256 - 144 \\ {r}^{2} = 112 \\ r \: = 10.89cm$