Question:
the locus of the centre of the circles such that the point (2,3) is the mid point of the chord 5x+2y=16 is?
Answer:
Let Chord be AB,
and, OD be perpedicular on AB, where, D is mid point of AB.
Then, slope of AB : 5x + 2y = 16 is -5/2
thus, slope of OD = 2/5
=> y = mx + c
=> c = y - mx = 3 - 2 x 2/5 = 11/5
Thus the required locus is, y = 2/5 x + 11/5 i.e., 2x - 5y + 11 = 0.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
