Question

prove that into words and circle the chords of the larger circle which touches the smaller circle is bisected at the point of contact​

Answer 1

Step-by-step explanation:

We are given two concentric circles C1 and C2 with centre O and a chord AB of the larger circle C1 which touches the smaller circle C2 at the point P (see Fig. 10.8). We need to prove that AP = BP. Let us join OP. Then, AB is a tangent to C2 at P and OP is its radius. Therefore, OP⟂AB (Since Tangent at any point of circle is perpendicular to the radius through point of contact) Now AB is a chord of the circle C1 and OP⟂AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord, Hence AP = BP