Question number 5 Q5. From (3, 4) chords are drawn to the circle x2 + y2 – 4x = 0. The locus of the mid points of the chords is (A) x2 + y2 – 5x – 4y + 6 = 0 (B) x2 + y2 +5x – 4y + 6 = 0 (C) x2 + y2 – 5x + 4y + 6 = 0 (B) x2 + y2 – 5x – 4y – 6 = 0 Share with your friends Share 3 Neha Sethi answered this Dear student We have,x2+y2-4x=0On comparing with the equation x2+y2+2gx+2fy+c=02fy=0⇒f=0 and 2gx=-4xg=-2and we know that centre is given by : -g,-f=--2,0=2,0In the figure O 2,0 is the centre of the circle , P is any point with coordinates 3,4 .Now we will find the locus of the mid points of the chord.Let the coordinates of M be h,k We know that line passing from the centre of the circle is ⊥ bisector of the chord⇒∠OMP=90°Let slope of OM=m1 and MP=m2and products of slope of OM and MP will be -1 as lines are ⊥⇒m1×m2=-1 ...1Slope of m1=0-k2-h and slope of m2=4-k3-hSo, putting in 1, we get0-k2-h4-k3-h=-14-kk=-13-hh-2k2-4k=3h-6-h2+2hk2-4k=5h-6-h2h2+k2+6-5h-4k=0x2+y2-5x-4y+6=0 On generalising is the required answer. Regards 5 View Full Answer
