Answered

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In the standard equation for a conic section Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, if B2 − 4AC < 0, the conic section in question is a circle if A = C , B = 0.TrueFalse

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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