Question

Prove that the perpendicular at the point of contact to the tangent to a circle through the centre. distances​

Answer 1

$\bf \red{Qᴜᴇsᴛiᴏɴ-:}$

⇝ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ᴛʜᴇ ᴘᴇʀᴘᴇɴᴅɪᴄᴜʟᴀʀ ᴀᴛ ᴛʜᴇ ᴘᴏɪɴᴛ ᴏꜰ ᴄᴏɴᴛᴀᴄᴛ ᴛᴏ ᴛʜᴇ ᴛᴀɴɢᴇɴᴛ ᴛᴏ ᴀ ᴄɪʀᴄʟᴇ ᴛʜʀᴏᴜɢʜ ᴛʜᴇ ᴄᴇɴᴛʀᴇ.

$\bf \red{ᴀɴsᴡᴇʀ-:}$

Let the centre of cirlce be O.

And AB the tangent point intersecting circle at point P.

And to prove that OP⊥AB.

As we know that tangent of the circle is perpendicular to radius at points of contact Hence

OP⊥AB

So, ∠OPB=90°(i)

Now lets assume some point X

Such that XP⊥AN

Hence ∠XPB=90°(ii)

From eq (i) & (ii)

∠OPB=∠XPB=90°

Which is possible only if line XP passes though O

Hence perpendicular to tangent passes though centre.

Answer 2

Step-by-step explanation:

${ \sf{ \color{maroon}{Question:}}}$

Prove that the perpendicular at the point of contact to the tangent to a circle through the centre.

${ \sf{ \color{maroon}{Required \: Solution :}}}$

Let, O is the centre of the given circle.

A tangent PR has been drawn touching the circle at point P.

Draw QP ⊥ RP at point P, such that point Q lies on the circle.

∠OPR = 90° (radius ⊥ tangent)

Also, ∠QPR = 90° (Given)

∴ ∠OPR = ∠QPR

Now, the above case is possible only when centre O lies on the line QP.

–Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

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