FIND THE EQUATION OF THE CIRCLE WHICH PASSES THROUGH 1. THE POINT (-2,1) AND IS TANGENT TO THE LINE 3X - 2Y - 6=0 AT THE POINT (4,3) Share with your friends Share 0 Rahul Raj answered this dear student normal to line 3x-2y-6=0 at (4,3) is2x+3y=17perpendicular bisector to chord joining (-2,1) and (4,3) passes through centremidpoint of (-2,1) and (4,3) is (1,2) and slope of chord is 1/3normal is y-2=-3(x-1)3x+y=5centre is intersection of 2x+3y=17 and 3x+y=5which is (-2/7,41/7)circle (x+27)2+(y-417)2=(4-27)2+(3-417)2(x+27)2+(y-417)2=107649 regards 2 View Full Answer Asha Jerein answered this this question won't come in exam 0