Question

The tangent at any point of a circle is perpendicular to the radius through the point of contact
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Answer 1

GIVEN:

• A circle and a tangent YX at a point P

TO PROVE:

• OP⊥ YX

CONSTRUCTION:

• Take a point Q other than P, on tangent YX. Join OQ.

PROOF:

Since, Q is a point on the tangent AB, other than the point P, so Q will be outside the circle.

☞ Let OQ intersect the circle at R.

$\rm{\longrightarrow OQ = OR + RQ }$

$\rm{\longrightarrow OQ > OR }$

$\rm{\longrightarrow OQ > OP }$ [OP = OR = Radius]

$\rm{\longrightarrow OP < OQ }$

Thus, OP is shorter than any other segment joining O to any point of YX.

But among all line segments, joining the point O to a point on YX, the shortest one is the perpendicular from O on YX.

Hence,

⠀⠀⠀ $\bf{ OP \perp YX}$

Answer 2

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

Then, OR < OQ ⠀ [a part is less than the whole] ...❶

But, OP = OR ⠀ [radii of the same circle whole] ...❷

∴ OR < OQ ⠀ [from equ ❶ and ❷]

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

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

REMARKS

(i) We conclude that any point on a circle, one and only one tangent can be drawn to the circle.

(ii) The line containing the radius through the point of contact is called the “normal” to the circle at point of the contact.