Question

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

Answer 1

Answer:

Construction : Take a point B, other than A, on the tangent l. ... 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.Feb 8, 2018

Step-by-step explanation:

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Answer 2

Figure of the question

$\mathtt{\bf{\underline{\red{Given:-}}}}$

• A circle C (0, r).

• A tangent l at point A touches circle at point A.

$\mathtt{\bf{\underline{\green{To\:Proof:-}}}}$

• OA ⊥ l

$\mathtt{\bf{\underline{\purple{Construction:-}}}}$

Take a point B on the tangent l & Join OB.

$\mathtt{\bf{\underline{\orange{Proof:-}}}}$

We know that, if the the line segment joins the radius to the tangent l . So, the perpendicular will become shortest as compared to l.

OA = OC (Radii of the same circle)

According to the figure,

OB = OC + BC.

∴ OB > OC

➝ OB > OA

➝ OA < OB

We know that any arbitrary point B on the tangent l.

So , we can say that OA is shorter line segment as compared to other line segments joining O to any point on l.

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

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