Question

the tangent of any point of a circle is perpendicular to the radius through the point of contact​

Answer 1

Answer:

Qᴜᴇsᴛɪᴏɴ :

➥ The tangents at any point of a circle is perpendicular to the radius through the point of contact.

Pʀᴏᴠᴇᴅ :

➥ OP ⊥ AB

Gɪᴠᴇɴ :

➤ A circle with center O and a tangent AB at a point P of the circle.

Tᴏ Pʀᴏᴠᴇ :

➤ OP ⊥ AB ?

Cᴏɴꜱᴛʀᴜᴄᴛɪᴏɴ :

➤ Take a point Q, other than P, on AB. Join OQ.

Pʀᴏᴏꜰ :

Q is a point on the tangent AB, other than the point of contact P.

∴ Q lies outside the circle

Let OQ intersect the circle at R

$\sf \: Then, OR < OQ \: \: \: \bigg[a \: part \: is \: less \: than \: the \: whole \bigg] ...❶$

$\sf \: But, OP = OR \: \: \: \bigg[radii \: of \: the \: same \: circle \: whole \bigg] ...❷$

$\sf∴ OR < OQ \: \: \: \: \: \: \: \bigg[from \: equ \: ❶ \: and \: ❷ \bigg]$

Thus, OP is shorter than any other line segment joining O to any point point of AB, other than P

In other words, OP is the shortest distance between the point O and the line AB

But, the shortest distance between a point and a line is the perpendicular distance

$:\implies \underline{\overline{\boxed{\purple{\bf{\:\:\therefore OP \perp AB \:\:}}}}}:$

Hence proved.