Question

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 `(c ,0)dot`

Answer 1

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`