Home
Class 12
MATHS
Find the locus of the midpoint of the ch...

Find the locus of the midpoint of the chords of the circle `x^2+y^2=a^2` which subtend a right angle at the point `(0,0)dot`

Text Solution

Verified by Experts

Let P(h,k) be the midpoint of a chord AB of the circle which subtends a right angle at C(c,0).

In right angles triangle ACB, we have
`PA=PC=PB`
`PC=sqrt((h-c)^(2)+k^(2))`
`PB=PA=sqrt(OA^(2)-OP^(2))`
`=sqrt(a^(2)-(h^(2)+k^(2)))`
Now , `PC=PA`
`:. sqrt((h-c)^(2)+k^(2))=sqrt(a^(2)-(h^(2)+k^(2)))`
`implies (h-c)^(2)+k^(2)=a^(2)-(h^(2)+k^(2))`
Therefore , the equation of required locus is `2(x^(2)+y^(2))-2cx+c^(2)-a^(2)=0`
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

Find the locus of the middle points of the chords of the parabola y^2=4a x which subtend a right angle at the vertex of the parabola.

The locus of the midpoint of a chord of the circle x^2+y^2=4 which subtends a right angle at the origins is (a) x+y=2 (b) x^2+y^2=1 (c) x^2+y^2=2 (d) x+y=1

Find the locus of the midpoint of the chords of circle x^(2)+y^(2)=a^(2) having fixed length l.

The locus of the midpoints of the chords of the circle x^2+y^2-a x-b y=0 which subtend a right angle at (a/2, b/2) is (a) a x+b y=0 (b) a x+b y=a^2=b^2 (c) x^2+y^2-a x-b y+(a^2+b^2)/8=0 (d) x^2+y^2-a x-b y-(a^2+b^2)/8=0

Find the locus of the midpoint of the chord of the circle x^2+y^2-2x-2y=0 , which makes an angle of 120^0 at the center.

Find the locus of the midpoint of normal chord of parabola y^2=4ax

Find the locus of the-mid points of the chords of the circle x^2 + y^2=16 , which are tangent to the hyperbola 9x^2-16y^2= 144

The locus of the mid-points of the chords of the circle of lines radiùs r which subtend an angle pi/4 at any point on the circumference of the circle is a concentric circle with radius equal to

The locus of the middle points of the focal chords of the parabola, y^2=4x is:

The equation of the locus of the mid-points of chords of the circle 4x^2 + 4y^2-12x + 4y +1= 0 that subtends an angle of at its centre is (2pi)/3 at its centre is x^2 + y^2-kx + y +31/16=0 then k is