Question

The equation of the chord joining two points P(x₁, y₁) and Q(x₂, y₂) on a circle S = 0 is S₁ + S₂ = S₁₂ and hence the equation of the tangent at (x₁, y₁) is S₁ = 0.

Answer 1

Let S ≡ x² + y² + 2gx + 2fy + c = 0. Since P(x₁, y₁) and  Q(x₂, y₂) lie on the circle (see Fig.), we have

Let C = (−g, −f ) (centre) and M = (x₁ + x₂/2 , y₁ + y₂/2 (the midpoint of (bar)AB ). From (bar) AB is perpendicular to (bar)CM so that the equation of the chord (bar)AB is

Therefore

That is, equation of the chord AB is S₁ + S₂ = S₁₂.

Since the tangent at P(x₁, y₁) to the circle is the limiting position of the chord (bar)PQ as Q approaches P along the circle , the equation of the tangent is

S₁ + S₂ = S₁₁ = 0 

Hence, S₁  ≡  xx₁ +  yy₁ + g(x + x₁) + f(y + y₁) + c = 0 is the tangent at (x₁, y₁).