If from any point `P` on the circle `x^2+y^2+2gx+2fy+c=0,` tangents are drawn to the circle `x^2+y^2+2gx+2fy+csin^2alpha+(g^2+f^2)cos^2alpha=0` , then find the angle between the tangents.

Explanation

Given circles are `x^(2)+y^(2)+2gx+2fy+c=0` (1)
and `x^(2)+y^(2)+2gx+2fy+c sin ^(2)alpha+(g^(2)+f^(2))cos^(2)alpha=0` (2)

Let `P(x_(1),y_(1))` be a point on the circle (1).
`:. x_(1)^(2)+y_(1)^(2)+2gx_(1)+2fy_(1)+c=0` (3)
The length of the tangent drawn from `P(x_(1),y_(1))` to the circle (2) is
`PQ=sqrt(x_(1)^(2)+y_(1)^(2)+2gx_(1)+2fy_(1)+csin^(2)alpha+(g+f^(2))cos^(2)alpha)`
`=sqrt(-c+csin^(2)alpha+(g^(2)+f^(2))cos^(2)alpha))` (Using (1))
`=(sqrt(g^(2)+f^(2)-c))cosalpha`
The radius of the circle (2) is
`CQ=(sqrt(g^(2)+f^(2)-csin^(2)alpha-(g^(2)+f^(2))cos^(2)alpha)`
`=(sqrt(g^(2)+f^(2)-c))sin alpha`
In `Delta CPQ`
`tan theta=(CQ)/(PQ)=(sqrt(g^(2)+f^(2)-c)sinalpha)/(sqrt(g^(2)+f^(2)-c)cosalpha)=tan alpha`
`:. theta= alpha`
Therefore, angle between tangents is `2 alpha`.