Math, asked by MASTERMIND0264, 6 months ago

prove that the tangents at any point of a circle is perpendicular to the radius through point of contact​

Answers

Answered by Anonymous
11

 \bold \red{given}

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

 \bold \blue{To prove}

OA ⊥ l

 \bold \purple{construction}

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

 \bold \pink{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.

 \bold \green{Hence}

Here, OA ⊥ l

Attachments:
Answered by ItZzMissKhushi
3

Answer:

given

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

Toprove

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.

Hence

Here, OA ⊥ l

Step-by-step explanation:

Hope this helps

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