Q. 8

# Prove that the pe

Let us consider a circle with center O and XY be a tangent

To prove : Perpendicular at the point of contact of the tangent to a circle passes through the center i.e. the radius OP XY

Proof :

Take a point Q on XY other than P and join OQ .

The point Q must lie outside the circle. (because if Q lies inside the circle, XY

will become a secant and not a tangent to the circle).

OQ is longer than the radius OP of the circle. That is,

OQ > OP.

Since this happens for every point on the line XY except the point P, OP is the

shortest of all the distances of the point O to the points of XY.

So OP is perpendicular to XY.

[As Out of all the line segments, drawn from a point to points of a line not passing through the point, the smallest is the perpendicular to the line.]

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