Question

1. Find the equation of tangent to the circle $x^2 + y^2 - 2ax = 0$ at the point $(a(1 + cos\alpha), asin\alpha)$.

Answer 1

The equation of the tangent is $x\cos\alpha + y\sin\alpha = a(1 + \cos\alpha)$.

Answer 2

The equation of the circle is $x^2 + y^2 - 2ax = 0$. The center of the circle is $(a, 0)$ and the radius is $a$. The given point is $P(a(1 + \cos\alpha), a\sin\alpha)$. Using the formula for the tangent to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at a point $(x_1, y_1)$, which is $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. For the given circle, $g = -a$, $f = 0$, $c = 0$. Substituting $(x_1, y_1) = (a(1 + \cos\alpha), a\sin\alpha)$: $x[a(1 + \cos\alpha)] + y[a\sin\alpha] - a[x + a(1 + \cos\alpha)] + 0 + 0 = 0$. Simplifying this equation leads to $x\cos\alpha + y\sin\alpha = a(1 + \cos\alpha)$.