A circle is drawn through the points (3, 2), (-1, -2) and (5, -4). Find the :
(a) coordinates of the centre of the circle ;
(b) radius of the circle ;
(c) equation of the circle.
Use the general circle \(x^2+y^2+2gx+2fy+c=0\), with centre \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\). Substitute the three points.
\((3,2)\): \(6g+4f+c=-13\) (i)
\((-1,-2)\): \(-2g-4f+c=-5\) (ii)
\((5,-4)\): \(10g-8f+c=-41\) (iii)
(i)\(-\)(ii): \(8g+8f=-8\Rightarrow g+f=-1\). (iii)\(-\)(i): \(4g-12f=-28\Rightarrow g-3f=-7\).
Solving: \(f=\tfrac32,\ g=-\tfrac52\). From (ii): \(5-6+c=-5\Rightarrow c=-4\).
(a) Centre \((-g,-f)=\left(\tfrac52,\,-\tfrac32\right)\).
(b) Radius \(=\sqrt{g^2+f^2-c}=\sqrt{\tfrac{25}{4}+\tfrac94+4}=\sqrt{\tfrac{50}{4}}=\dfrac{5\sqrt2}{2}\approx3.54\).
(c) Equation:
\[x^2+y^2-5x+3y-4=0\]
(Check with \((3,2)\): \(9+4-15+6-4=0\).)