IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

deep01

From the origion chords r drawn to the cirle (x-1)²+y²=1. the equation of the locus of the mid-points of these chords is??

pls also explain ur answer

ankurgupta91

let the mid-point of these chords be P(x1,y1)
as it is the mid-point of line joining the origin with the other end of circle
so the co-ordinates of other end will be(2x1,2y1)
nw this point satisfy the equation of circle
thus , (2x1-1)^2 + (2y1)^2 =1
4x1^2-4x1+4y1^2=0
thus locus is given by
4x^2 - 4x +4y^2 =0

truly

The chords drawn from the origin would have the equation y = mx (1)

the equation of the circle is (x-1)²+y_²= 1 (2)

So finding the points of intersection using (1) and (2) equations

we get x=0, y=0 & x=2/(m²+1) , y= 2m/(m²+1)

therefore the midpoint (h,k) is

h=1/(m²+1) , k= m/(m²+1)

Therefore k/h = m

Using in h=1/(m²+1)

=> h(k²/h² + 1) = 1

k²+h² - h =0

therefore locus of (h,k) is

y²+x² - x=0

(x-1/2)² + y² =1/4

vinkupaan

The chords drawn from the origin would have the equation y = mx (1)

the equation of the circle is (x-1)2+y2 = 1 (2)

So finding the points of intersection using (1) and (2) equations

we get x=0, y=0 & x=2/(m2+1) , y= 2m/(m2+1)

therefore the midpoint (h,k) is

h=1/(m2+1) , k= m/(m2+1)

Therefore k/h = m

Using in h=1/(m2+1)

=> h(k2/h2 + 1) = 1

k2+h2 - h =0

therefore locus of (h,k) is

y2+x2 - x=0

(x-1/2)2 + y2 =1/4

vinkupaan

The chords drawn from the origin would have the equation y = mx (1)

the equation of the circle is (x-1)²+y_²= 1 (2)

So finding the points of intersection using (1) and (2) equations

we get x=0, y=0 & x=2/(m²+1) , y= 2m/(m²+1)

therefore the midpoint (h,k) is

h=1/(m²+1) , k= m/(m²+1)

Therefore k/h = m

Using in h=1/(m²+1)

=> h(k²/h² + 1) = 1

k²+h² - h =0

therefore locus of (h,k) is

y²+x² - x=0

(x-1/2)² + y² =1/4

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