What is the slope of the chord of contact of the point (6, -4) to the circle x2+y2−2x−2y+1=0
1
The given circle can be represented by the circle below. RS is the chord of the contact and OP is the line of the segmant joining center of the circle and the point P.
ΔPRS is symmetrically cut by ¯¯¯¯¯¯¯¯OP and since ¯¯¯¯¯¯¯¯OP is perpendicular to ¯¯¯¯¯¯¯¯RS , the product of their slopes would be -1. if m is slope of ¯¯¯¯¯¯¯¯RS,
m x slope of OP =-1
mx[−4−16−1]=−1
mx−55=−1
∴m=1