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Answer:
[tex](x-5)^2+(y-10)^2=65[/tex]
Step-by-step explanation:
The general equation of a circle is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General equation of a circle}}\\\\(x-h)^2+(y-k)^2=r^2\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\textsf{$(h, k)$ is the center.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius.}\end{array}}[/tex]
Given that the circle passes through the points (1, 3), (-2, 6), and (4, 2), we can substitute the coordinates of each point into the general equation to create a system of three equations:
[tex](1-h)^2+(3-k)^2=r^2\\\\(-2-h)^2+(6-k)^2=r^2\\\\(4-h)^2+(2-k)^2=r^2[/tex]
Set the first equation equal to the second equation:
[tex](1-h)^2+(3-k)^2=(-2-h)^2+(6-k)^2[/tex]
Solve for k:
[tex]1-2h+h^2+9-6k+k^2=4+4h+h^2+36-12k+k^2 \\\\ -2h+10-6k=4h+40-12k \\\\ 6k=6h+30\\\\k=h+5[/tex]
Set the second equation equal to the third equation:
[tex](-2-h)^2+(6-k)^2=(4-h)^2+(2-k)^2[/tex]
Solve for k:
[tex]4+4h+h^2+36-12k+k^2=16-8h+h^2+4-4k+k^2 \\\\4h+40-12k=20-8h-4k \\\\-8k=-12h-20\\\\8k=12h+20\\\\2k=3h+5\\\\k=\dfrac{3h+5}{2}[/tex]
Set the two equations for k equal to each other, and solve for h:
[tex]\dfrac{3h+5}{2}=h+5 \\\\3h+5=2h+10\\\\h = 5[/tex]
Substitute h = 5 into one of the equations for k, and solve for k:
[tex]k = h + 5 \\\\ k = 5 + 5 \\\\ k = 10[/tex]
Now, substitute h = 5 and k = 10 into one of the equations and solve for r²:
[tex](1-h)^2+(3-k)^2=r^2\\\\(1-5)^2+(3-10)^2=r^2 \\\\(-4)^2+(-7)^2=r^2\\\\16+49=r^2\\\\r^2=65[/tex]
Finally, substitute h = 5, k = 10 and r² = 65 into the general equation of a circle:
[tex](x-5)^2+(y-10)^2=65[/tex]
Therefore, the equation of the circle that passes through the points (1, 3), (-2, 6), and (4, 2) is:
[tex]\Large\boxed{\boxed{(x-5)^2+(y-10)^2=65}}[/tex]
This circle has a center located at point (5, 10) and a radius of [tex]\sqrt{65}[/tex].
