Question

26. Prove that radius of a circle is perpendicular to the tangent at the point of contact​

Answer 1

Answer:

Given- a circle with centre O with tangent XY at the point of contact p

To prove - OP perpendicular at XY

Proof - Let Q be point on XY connect OQ

suppose it touches the circle at R

Hence

OQ is greater than OR

OQ is greater than OP

as OP or OR are radius

same will be the case with all other points on circle, Hence OPis the smallest line that connects XY and smallest line is perpendicular.

Therefore OP

perpendicular at XY

Explanation:

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Answer 2

Given : A circle C (0, r) and a tangent l at point A.

To prove : OA ⊥ l

Construction : Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.

Proof: We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.

OA = OC  (Radius of the same circle)

Now, OB = OC + BC.

∴ OB > OC

⇒ OB > OA

⇒ OA < OB

B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l.

Here, OA ⊥ l