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Sagot :
For this problem we start from the equation given and susbtitute the conditions:
[tex]\begin{gathered} Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \\ WenowusethatA=CandB=0andC>0(thiscomesfromB^2-4AC<0) \\ Cx^2+Cy^2+Dx+Ey+F=0 \\ \frac{C}{C}x^2+\frac{^{}D}{C}x+\frac{C}{C}y^2+\frac{E}{C}y=-\frac{F}{C} \end{gathered}[/tex]Now we complete the perfect square trinomials:
[tex]\begin{gathered} x^2+\frac{D}{C}x+(\frac{D}{2C})^2+y^2+\frac{E}{C}y+(\frac{E}{2C})^2=-\frac{F}{C}+(\frac{D}{2C})^2+(\frac{E}{2C})^2 \\ (x+\frac{D}{2C})^2+(y+\frac{E}{2C})^2=-\frac{F}{C}+(\frac{D}{2C})^2+(\frac{E}{2C})^2=\text{ constant } \end{gathered}[/tex]Now, we have arrived to the general form of an equation of a circle.
Answer: True.
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