Tangent to two circles with parametric coordinates (2+2cos θ , 2 sin θ)?

Question

I was wondering if someone could help solve this question,

Show that the point with coordinate$ ( 2 + 2cos (θ) , 2 sin (θ))$ lie on the circle $x^2 + y^2 = 4x$ and obtain the equation of the tangent to the circle at this point.

The tangent at the points A and B on this circle touch the circle $x^2+y^2=1$ at the points C and D. Find the coordinate of the points of intersection of these tangents, and obtain the equation of the circle through the points A B C D.

the answers are:

$x cos θ + y sin θ = 2 + 2 cos θ $
(-2,0)
$x^2+y^2=2x+2$

I can get the equation for the tangent quite easily but can't get the rest, I run into some really messy algebra, by inserting x,in the equation of the tangent, into the formula for the circle, to find the points of intersection. The algebra is horrid and leads to no viable answer.

Thanks for any help.

Answer 1

$ x^2+y^2=4x $ and $ (x-2)^2+y^2=2^2 $ represent the same circle! ( being tangent to y-axis at the origin).

Answer 2

You can avoid the messy algebra via another method. The parametric equation for the smaller circle is x=cosθ, y=sinθ and you can use this as the coordinates on the smaller circle. If you sketch the graph for both circles and draw the tangent line that touches both circles at A on the bigger circle and C on the smaller circle you will notice the parameter or angle θ subtended for point A on the bigger circle and point C on the smaller circle are the same because the line from center of the circle to the point A and C respectively are perpendicular to the tangent line. enter image description here

The gradient of the tangent line is given by the derivative of the given point, m = -cosθ/sinθ, which is the same if you work out the gradient between points A and C, which is 2sinθ-sinθ/2+2cosθ-cosθ=-cosθ/sinθ. Solving this equation to get the values of θ you can proceed to get the equation of the other tangent and obtain the intersection point...