Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Using the above, do the following:
In fig., O is the center of the two concentric circles. AB is a chord of the larger circle touching the smaller circle at C. Prove that AC = BC
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Using the above, do the following:
In fig., O is the center of the two concentric circles. AB is a chord of the larger circle touching the smaller circle at C. Prove that AC = BC
Take a point Q on XY other than P and join OQ.
The point Q must lie outside the circle.
Therefore, 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.
Join OC.
Since OP is a radius of smaller circle and ACB is a tangent to it at the point P, so OC perpendicular to AB.
But the perpendicular from the center to a chord, bisects the chord.
AC = CB
Hence, AB is bisected at the point C.
Take a point Q on XY other than P and join OQ.
The point Q must lie outside the circle.
Therefore, 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.
Join OC.
Since OP is a radius of smaller circle and ACB is a tangent to it at the point P, so OC perpendicular to AB.
But the perpendicular from the center to a chord, bisects the chord.
AC = CB
Hence, AB is bisected at the point C.
