The points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle.
circle passes through the points A, B, C.
therefore, the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1)
Substituting A (5, 5) in (1), we get,
\[\begin{array}{*{35}{l}}
{{5}^{2}}~+\text{ }{{5}^{2}}~+\text{ }2a\left( 5 \right)\text{ }+\text{ }2b\left( 5 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
25\text{ }+\text{ }25\text{ }+\text{ }10a\text{ }+\text{ }10b\text{ }+\text{ }c\text{ }=\text{ }0 \\
10a\text{ }+\text{ }10b\text{ }+\text{ }c\text{ }+\text{ }50\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right) \\
\end{array}\]
Substituting the points B (6, 4) in equation (1), we get,
\[\begin{array}{*{35}{l}}
{{6}^{2}}~+\text{ }{{4}^{2}}~+\text{ }2a\left( 6 \right)\text{ }+\text{ }2b\left( 4 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
36\text{ }+\text{ }16\text{ }+\text{ }12a\text{ }+\text{ }8b\text{ }+\text{ }c\text{ }=\text{ }0 \\
12a\text{ }+\text{ }8b\text{ }+\text{ }c\text{ }+\text{ }52\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right) \\
\end{array}\]
Substitute the point C (-2, 4) in equation (1), we get,
\[\begin{array}{*{35}{l}}
{{\left( -2 \right)}^{2}}~+\text{ }{{4}^{2}}~+\text{ }2a\left( -2 \right)\text{ }+\text{ }2b\left( 4 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
4\text{ }+\text{ }16\text{ }-\text{ }4a\text{ }+\text{ }8b\text{ }+\text{ }c\text{ }=\text{ }0 \\
20\text{ }-\text{ }4a\text{ }+\text{ }8b\text{ }+\text{ }c\text{ }=\text{ }0 \\
4a\text{ }-\text{ }8b\text{ }\text{ }c\text{ }\text{ }20\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\
\end{array}\]
Upon simplifying equations (2), (3) and (4) we get,
a = – 2, b = – 1 and c = – 20
Now by substituting the values of a, b, c in equation (1), we get
\[\begin{array}{*{35}{l}}
{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( -\text{ }2 \right)x\text{ }+\text{ }2\left( -\text{ }1 \right)y\text{ }\text{ }20\text{ }=\text{ }0 \\
{{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }4x\text{ }-\text{ }2y\text{ }\text{ }20\text{ }=\text{ }0\text{ }\ldots ..\text{ }\left( 5 \right) \\
\end{array}\]
Substituting D (7, 1) in equation (5) we get,
\[\begin{array}{*{35}{l}}
{{7}^{2}}~+\text{ }{{1}^{2}}~-\text{ }4\left( 7 \right)\text{ }-\text{ }2\left( 1 \right)\text{ }-\text{ }20 \\
49\text{ }+\text{ }1\text{ }-\text{ }28\text{ }-\text{ }2\text{ }-\text{ }20 \\
0 \\
\end{array}\]
∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) lie on a circle.
Now let us find the centre and the radius.
We know that for a circle x2 + y2 + 2ax + 2by + c = 0,
Centre = (-a, -b)
Radius = √(a2 + b2 – c)
Comparing equation (5) with equation (1), we get
Centre = [-(-4)/2, -(-2)/2)]
= (2, 1)
Radius = √(22 + 12 – (-20))
= √(25)
= 5
∴ The centre and radius of the circle is (2, 1) and 5.