IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

dinesh_ddt

Find the locus of mid points of the chords of the circle 4x^2+4y^2-12x+4y+1=0 that subtends an angle of 120 degrees at the centre ?

yahiyafirdous

Let AB is a chord which makes 120⁰

Centre of the circle is O(3/2, -1/2)

C(x_1,y₁) is mid point of chord.

Now OC is perpendicular to AB

Sin(AngleOAC)=

[(x₁- 3/2)² + (y₁+1/2)²] / OA

Sin30=

[(x₁- 3/2)² + (y₁+1/2)²] / (3/2)

Required locus is:

(x- 3/2)² + (y+1/2)²=(3/4)²

^{Hey man do not gorget to rate me}

sudeep.kumar

Firdous is correct, and if you want to solve this problem anlytically.. go fot that solution...

But geometrically it is a very simple question and you should be able to write the equation of the locus in one step.
The logic is...
In a circle, equal chords make equal angles at the centre, and they are equidistant from the centre. So, all the mid points will be at same fixed distance from the centre of the given circle, and that distance is r cos (

/2) and

here is 120^o, ie the locus is always at a dist of r/2 from the centre.

Now you can easily write the equation of the circle, give its centre being same as that of the given one and radius half of that... and thats ur locus.

Ask & Discuss Questions with Community & Experts