Question

Prove that, the tangent to a circle at any point on it is perpendicular to the radius passes
through the point of contact.
➡️ Wrong Answer will be reported ​

Answer 1

Given:

✭ A circle with center O

✭ AB is a tangent drawn through P

To Prove:

◈ OP ⟂ AB

Solution:

Theorem

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

Construction

• Draw OQ outside the circle

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Proof

»» The Poin Q is outside the circle

»» As it is outside the circle it should be longer than the radius

»» So OQ longer than the radius (OQ > OP)

• This happens for any point on AB except OP which the shortest distance of all the other distance from the point O to the line AB

∴ OP ⟂ AB

Hence Proved!!

[Diagram has been attached]

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☯ Know More

⪼ The length of the tangents drawn from an external point to a circle are equal

⪼ The tangents at the extremeties of any chord make equal agles with the chord

⪼ The angle made by the chord and tangent is equal to the angle made in the alternate segments

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Answer 2

Given:

AB is a tangent touching the circle at point A.

TO Prove:

The tangent to a circle at any point on it is perpendicular to the radius passes through the point of contact.

Construction:

Join OB touching the circle at C.

Step by step proof:

OA=OC (Radii of circle)

Now OB=OC+BC

∴OB>OC (OC being radius and B any point on tangent)

⇒OA<OB

B is an arbitrary point on the tangent.

Thus, OA is shorter than any other line segment joining O to any

point on tangent.

Shortest distance of a point from a given line is the perpendicular distance from that line.

Hence, the tangent at any point of circle is perpendicular to the radius.