Centre, radius and position of a point

Problem

Given $\gamma:\ x^2 + y^2 - 6x + 2y + 6 = 0$, determine the centre and radius; check whether the point $A(1;1)$ is internal, external or on the circle.

Solution

From the standard form we read $a = -6$, $b = 2$, $c = 6$. Hence $C\left(-\tfrac{a}{2};\ -\tfrac{b}{2}\right) = C(3;-1), \qquad R^2 = \tfrac{a^2}{4} + \tfrac{b^2}{4} - c = 9 + 1 - 6 = 4,$ that is $\boxed{C(3;-1),\ R = 2.}$

For the position of $A(1;1)$ we compute its distance from the centre: $d(A,C) = \sqrt{(1-3)^2 + (1+1)^2} = \sqrt{4 + 4} = 2\sqrt{2} \approx 2{,}83.$ Since $d(A,C) = 2\sqrt{2} > 2 = R$, the point $A$ is external to the circle.

Links

Topics: Circonferenza analitica
Concepts: Centro · Distanza tra punti · Equazione circonferenza · Forma standard · Piano cartesiano · Punto interno esterno · Raggio
Methods: Circonferenza equazione
Skills: Geometria analitica
Exercise types: Problema geometrico