Problems from Circumference that goes through 3 given points

Find the circumference that goes through points $$a=(2,0)$$, $$b=(2,3)$$ and $$c=(1,3)$$.

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Development:

We replace them in the general equation of the circumference $$x^2+y^2+Ax+By+C=0$$. Thereby we obtain:

$$$\left\{\begin{array}{c} 2^2+0+2A+0+C=0 \\ 2^2+3^2+2A+3B+C=0 \\ 1^2+3^2+A+3B+C=0 \end{array}\right\} \Rightarrow \left\{\begin{array}{c} 4+2A=-C \\ 13+2A+3B+C=0 \\ 10+A+3B+C=0 \end{array}\right.$$$ We solve the system and get: $$$\left\{\begin{array}{c} A=-3 \\ B=-3 \\ C=2 \end{array}\right.$$$ And we have the equation $$$x^2+y^2-3x-3y+2=0$$$

Solution:

$$x^2+y^2-3x-3y+2=0$$

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