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

PA is the ten gent to the circle at point A

PA = 90* [ radius is always perpendicular to the tengent ]

In trg PAO+trg PAO = 90* [ form ¡ eq]

po^2= (12)^2+ AO^2

therefore AO^2 = (16)^2-(12)^2

AO^2= 256 - 144

AO^2 = UNDERROOT 112

AO = 10.59 approx.

hence the radius of the circle is 10.59...