Question

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the center​

Answer 1

let angle qpb=90degree_(1)

as op perpendicular ab

angle opb = 90 degree_(2)

from 1 and 2

angle qpb= angle opb

which is not possible

therefore, our supposition is wrong hence perpendicular at a point of contact ti the tangent to a circle passes through the centre

Answer 2

Solution:

First, draw a circle with center O and draw a tangent AB which touches the radius of the circle at point P.

To Pr oof: PQ passes through point O.

Now, let us consider that PQ doesn’t pass through point O. Also, draw a CD parallel to AB through O. Here, CD is a straight line and AB is the tangent.

From the above diagram, PQ intersects CD and AB at R and P respectively.

AS, CD ∥ AB,

Here, the line segment PQ is the line of intersection.

Now angles ORP and RPA are equal as they are alternate interior angles

So, ∠ORP = ∠RPA

And,

∠RPA = 90° (Since, PQ is perpendicular to AB)

∠ORP = 90°

Now, ∠ROP+∠OPA = 180° (Since they are co-interior angles)

∠ROP+90° = 180°

∠ROP = 90°

Now, it is seen that the △ORP has two right angles which are ∠ORP and ∠ROP. Since this condition is impossible, it can be said the supposition we took is wrong.