Question

prove that the perpendicular at the point of contact to the tangent to a circle passes througj the centre.​

Answer 1

Answer:

ANSWER

Given a circle with center O and AB the tangent intersecting circle at point P

and prove that OP⊥AB

We know that tangent of the circle is perpendicular to radius at points of contact Hence

OP⊥AB

So, ∠OPB=90

o

..........(i)

Now lets assume some point X

Such that XP⊥AN

Hence ∠XPB=90

o

.........(ii)

From eq (i) & (ii)

∠OPB=∠XPB=90

o

Which is possible only if line XP passes though O

Hence perpendicular to tangent passes though centre

HOPE. ITS , HELP

Answer 2

Step-by-step explanation:

Given

• A tangent is perpendicular to the point of contact.

To Prove:

• Perpendicular at point of contact to the tangent passes through centre .

Proof: Let in a circle with centre O. We have

• PM = tangent

• P = point of contact

Let assume that PQ ⟂ PM , where Q lies on the circle.

• ∠QPM = 90°

If possible let again assume that PQ not passes through the centre of circle i.e O.

Do a construction here : Join PO and produce it to meet the circle at R. Now we have

• PO = radius through the point of contact.

• PO ⟂ PM { we know that the tangent is perpendicular to the radius though the point of contact }

So, ∠OPM = 90° or ∠RPM = 90°

Thus, we have ∠QPM = ∠RPM = 90° {from above}

This is is possible only if P , Q and R are collinear , but a straight line cuts a circle in at the most two points.

∴ Points Q and R coincide.

So, we conclude that PQ passes through the centre of circle O.

Hence, the perpendicular at the point of contact to the tangent passes through the centre.

$\large\bold{\texttt {Proved }}$