Question

prove that the perpendicular at the center of point of contact to the tangent to a circle passes through the center

Answer 1

$\huge\star{\purple{\underline{\underline{\mathbb{EXPLANATION:}}}}}$

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First, draw a circle with center O and draw a tangent AB which touches the radius of the circle at point P.

To Proof: PQ passes through point O.

Now, let us consider that PQ doesn’t pass through point O.

Also, draw a CD parallel to AB through O. Here, CD is a straight line and AB is the tangent.

From the above diagram, PQ intersects CD and AB at R and P respectively.

AS, CD | | AB,

Here, the line segment PQ is the line of intersection.

Now angles ORP and RPA are equal as they are alternate interior angles

So, ∠ORP = ∠RPA

And,

∠RPA = 90° (Since, PQ is perpendicular to AB)

⇒ ∠ORP = 90°

Now, ∠ROP + ∠OPA = 180° (Since they are co-interior angles)

⇒∠ROP + 90° = 180°

⇒∠ROP = 90°

Now, it is seen that the △ORP has two right angles which are ∠ORP and ∠ROP.

Answer 2

$\huge\mathcal\colorbox{purple}{{\color{red}{ANSWER☆~~~}}}$

ꜰɪʀꜱᴛ, ᴅʀᴀᴡ ᴀ ᴄɪʀᴄʟᴇ ᴡɪᴛʜ ᴄᴇɴᴛᴇʀ ᴏ ᴀɴᴅ ᴅʀᴀᴡ ᴀ ᴛᴀɴɢᴇɴᴛ ᴀʙ ᴡʜɪᴄʜ ᴛᴏᴜᴄʜᴇꜱ ᴛʜᴇ ʀᴀᴅɪᴜꜱ ᴏꜰ ᴛʜᴇ ᴄɪʀᴄʟᴇ ᴀᴛ ᴘᴏɪɴᴛ ᴘ.

ᴛᴏ ᴘʀᴏᴏꜰ: ᴘQ ᴘᴀꜱꜱᴇꜱ ᴛʜʀᴏᴜɢʜ ᴘᴏɪɴᴛ ᴏ.

ɴᴏᴡ, ʟᴇᴛ ᴜꜱ ᴄᴏɴꜱɪᴅᴇʀ ᴛʜᴀᴛ ᴘQ ᴅᴏᴇꜱɴ’ᴛ ᴘᴀꜱꜱ ᴛʜʀᴏᴜɢʜ ᴘᴏɪɴᴛ ᴏ. ᴀʟꜱᴏ, ᴅʀᴀᴡ ᴀ ᴄᴅ ᴘᴀʀᴀʟʟᴇʟ ᴛᴏ ᴀʙ ᴛʜʀᴏᴜɢʜ ᴏ. ʜᴇʀᴇ, ᴄᴅ ɪꜱ ᴀ ꜱᴛʀᴀɪɢʜᴛ ʟɪɴᴇ ᴀɴᴅ ᴀʙ ɪꜱ ᴛʜᴇ ᴛᴀɴɢᴇɴᴛ. ʀᴇꜰᴇʀ ᴛʜᴇ ᴅɪᴀɢʀᴀᴍ ɴᴏᴡ.

ꜰʀᴏᴍ ᴛʜᴇ ᴀʙᴏᴠᴇ ᴅɪᴀɢʀᴀᴍ, ᴘQ ɪɴᴛᴇʀꜱᴇᴄᴛꜱ ᴄᴅ ᴀɴᴅ ᴀʙ ᴀᴛ ʀ ᴀɴᴅ ᴘ ʀᴇꜱᴘᴇᴄᴛɪᴠᴇʟʏ.

ᴀꜱ, ᴄᴅ ∥ ᴀʙ,

ʜᴇʀᴇ, ᴛʜᴇ ʟɪɴᴇ ꜱᴇɢᴍᴇɴᴛ ᴘQ ɪꜱ ᴛʜᴇ ʟɪɴᴇ ᴏꜰ ɪɴᴛᴇʀꜱᴇᴄᴛɪᴏɴ.

ɴᴏᴡ ᴀɴɢʟᴇꜱ ᴏʀᴘ ᴀɴᴅ ʀᴘᴀ ᴀʀᴇ ᴇQᴜᴀʟ ᴀꜱ ᴛʜᴇʏ ᴀʀᴇ ᴀʟᴛᴇʀɴᴀᴛᴇ ɪɴᴛᴇʀɪᴏʀ ᴀɴɢʟᴇꜱ

ꜱᴏ, ∠ᴏʀᴘ = ∠ʀᴘᴀ

ᴀɴᴅ,

∠ʀᴘᴀ = 90° (ꜱɪɴᴄᴇ, ᴘQ ɪꜱ ᴘᴇʀᴘᴇɴᴅɪᴄᴜʟᴀʀ ᴛᴏ ᴀʙ)

∠ᴏʀᴘ = 90°

ɴᴏᴡ, ∠ʀᴏᴘ+∠ᴏᴘᴀ = 180° (ꜱɪɴᴄᴇ ᴛʜᴇʏ ᴀʀᴇ ᴄᴏ-ɪɴᴛᴇʀɪᴏʀ ᴀɴɢʟᴇꜱ)

∠ʀᴏᴘ+90° = 180°

∠ʀᴏᴘ = 90°

ɴᴏᴡ, ɪᴛ ɪꜱ ꜱᴇᴇɴ ᴛʜᴀᴛ ᴛʜᴇ △ᴏʀᴘ ʜᴀꜱ ᴛᴡᴏ ʀɪɢʜᴛ ᴀɴɢʟᴇꜱ ᴡʜɪᴄʜ ᴀʀᴇ ∠ᴏʀᴘ ᴀɴᴅ ∠ʀᴏᴘ. ꜱɪɴᴄᴇ ᴛʜɪꜱ ᴄᴏɴᴅɪᴛɪᴏɴ ɪꜱ ɪᴍᴘᴏꜱꜱɪʙʟᴇ, ɪᴛ ᴄᴀɴ ʙᴇ ꜱᴀɪᴅ ᴛʜᴇ ꜱᴜᴘᴘᴏꜱɪᴛɪᴏɴ ᴡᴇ ᴛᴏᴏᴋ ɪꜱ ᴡʀᴏɴɢ.