Question

from the point of the circle x^2+y^2+2gx+2fy+c=0
drawn to the circle x^2+y^2+2gx+2fy+c sin^2α +(g^2+f^2)cos^2α=0 prove that the angle between them is 2α

Answer 1

Looks like this will involve a whole lot of algebra.

If point (a,b) is on the first circle it will satisfy that equation.

I forget the formula in coordinate geometry for equation of tangents to circle from an external point but it is something like S * S1 = T^2, I think
where S is the LHS of the equation of the second circle
S1 is the LHS that you get when you substitute (a,b) into equation (2)
T is the equation of a pair of tangents which for the second circle will look like this:
ax + by + g(x + a) + f(y + b) + csin^2α +(g^2+f^2)cos^2α
....
There may be an easier solution than solving the above but IDK.

Answer 2

I am assuming from the point on the first circle, two tangents are drawn to the second circle. The word "tangent" is missing in the problem.

This may be a slightly easier than the one above:

Circle 1: \( x^2+y^2+2gx+2fy+c = 0 \)
Center at (-g, -f); radius = \( \sqrt{g^2 + f^2 - c} \)

Circle 2: \( x^2+y^2+2gx+2fy+c\sin^2\alpha +(g^2+f^2)\cos^2\alpha = 0 \)
Center at (-g, -f); radius = \( \sqrt{g^2 + f^2 - c\sin^2\alpha -(g^2+f^2)\cos^2\alpha} \)

Let (p,q) be a point on circle 1 from which a tangent \( y = mx + b \) is drawn to circle 2.
Since (p,q) lies on circle 1, it must satisfy that equation:
\( p^2+q^2+2gp+2fq+c = 0 \) ---- (1)
Since (p,q) lies on the tangent,
\(q = mp + b\)
\(b = q - mp\)
\(y = mx + q - mp \)
\(mx - y + q - mp = 0\) ---- (2)

Distance of center of circle, \((-g,-f)\), to the line \(mx - y + q - mp = 0\) is:
\(\Large \frac {-mg + f + q - mp}{\sqrt{m^2+1}} \)
which should equal the radius of the second circle:
\(\Large \frac {-mg + f - mp}{\sqrt{m^2+1}} \) = \( \sqrt{g^2 + f^2 - c\sin^2\alpha -(g^2+f^2)\cos^2\alpha} \)
Square both sides, solve the quadratic equation in m. Let \(m_1, m_2\) be the roots.

The angle \(\theta\) between the tangents = \(\theta_2 - \theta_1 \) where \(\theta_1\) and \(\theta_2\) are the angles made by the two tangents with the x-axis.

\(\tan(\theta) = \tan(\theta_2 - \theta_1) = \Large \frac{\tan\theta_2 - \tan\theta_1}{1+\tan\theta_1\tan\theta_2} = \frac{m_2 - m_1}{1+m_1m_2} \)

If you prove \(\tan\theta = \tan2\alpha \) then you have solved the problem.