Find the locus of center of circle of radius 2 units, if intercept cut on the x-axis is twice of intercept cut on the y-axis by the circle.

Answer

`4x^(2)-y^(2)=12`

Explanation

Let the circle be `x^(2)+y^(2)+2gx+2fy+c=0`
Therefore , the intercept on the x-axis is `2 sqrt(g^(2)-c)`
Also, the intercept on the y-axis is `2 sqrt(f^(2)-c)`
According to the question,
`sqrt(g^(2)-c)=2sqrt(f^(2)-c)`
or `g^(2)-4f^(2)= -3c` (1)
Given, radius `=2`
`:. g^(2)+f^(2)-c=4` (2)
Eliminating c from (1) and (2), we get
`4g^(2)-f^(2)=12`
Therfore, the locus is `4x^(2)-y^(2)=12`