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