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

$\bf \huge{ \blue{ \underline{Question}}}$

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

$\bf \huge { \blue{ \underline{Answer}}}$

$\bf \red{ \underline{Given}}$

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

$\bf \red{ \underline{To \: \: Prove}}$

• OA | l

$\bf \red{ \underline{Construction}}$

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

$\bf \red{ \underline{Proof}}$

we know that,

Among all line segment joining the point O to a point on l , the perpendicular is shorter to l.

$\tt \: OA = OC ( Radius \: \: of \: \: the \: \: same \: \: circle)$

Now,

$\tt \: OB = OC +BC \\ \\ </strong></p><p></p><p><strong> \tt \: \therefore OB > OC \\ \\ \tt\implies OB> OA</strong><strong> </strong>\\ \\ \tt\implies OA>OB <strong> </strong> <strong>$

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 $\perp$ l

Answer 2

Step-by-step explanation:

Given

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

\bf \red{ \underline{To \: \: Prove}}

ToProve

OA | l

\bf \red{ \underline{Construction}}

Construction

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

\bf \red{ \underline{Proof}}

Proof

we know that,

Among all line segment joining the point O to a point on l , the perpendicular is shorter to l.

\tt \: OA = OC ( Radius \: \: of \: \: the \: \: same \: \: circle)OA=OC(Radiusofthesamecircle)

Now,

\begin{gathered}\tt \: OB = OC +BC \\ \\ \tt \: \therefore OB > OC \\ \\ \tt\implies OB > OA \\ \\ \tt\implies OA > OB\end{gathered}

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.