Tangents drawn from point P to the circle `x^(2)+y^(2)=16` make the angles `theta_(1)` and `theta_(2)` with positive x-axis. Find the locus of point P such that `(tan theta_(1)-tan theta_(2))=c` ( constant) .

Explanation

Let point P be (h,k).
Equation of tangent to circle `x^(2)+y^(2)=16` having slope m is
`y=mx+4sqrt(1+m^(2))`
If tangent is passing through P, then
`k=mh+4sqrt(1+m^(2))`
`implies (k-mh)^(2)=16+16m^(2)`
`implies (h^(2)-16)m^(2)-2hm+k^(2)-16=0`
Roots of this equation of `m_(1)` and `m_(2)`, where `m_(1)= tan theta_(1)` and `m_(2)=tan theta_(2)`.
Sum of roots, `m_(1)+m_(2)=(2hk)/(h^(2)-16)`
Product of roots, `m_(1)m_(2)=(k^(2)-16)/(h^(2)-16)`
Given that `(tan theta_(1)-tan theta_(2))=c`
`implies m_(1)-m_(2)=c`
`implies(m_(1)+m_(2))^(2)-4m_(1)m_(2)=c^(2)`
`=((2hk)/(h^(2)-16))^(2)-4(k^(2)-16)/(h^(2)-16)=c^(2)`
`implies 4h^(2)k^(2)-4(k^(2)-16)(h^(2)-16)=c^(2)(h^(2)-16)^(2)`
`implies 4x^(2)y^(2)-4(x^(2)-16)(x^(2)-16)=c^(2)(x^(2)-16)`
This is the required locus.