Question

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

Answer 1

For this problem we start from the equation given and susbtitute the conditions:

`\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 \\ (C)/(C)x^2+\frac{^{}D}{C}x+(C)/(C)y^2+(E)/(C)y=-(F)/(C) \end{gathered}`

Now we complete the perfect square trinomials:

`\begin{gathered} x^2+(D)/(C)x+((D)/(2C))^2+y^2+(E)/(C)y+((E)/(2C))^2=-(F)/(C)+((D)/(2C))^2+((E)/(2C))^2 \\ (x+(D)/(2C))^2+(y+(E)/(2C))^2=-(F)/(C)+((D)/(2C))^2+((E)/(2C))^2=\text{ constant } \end{gathered}`

Now, we have arrived to the general form of an equation of a circle.

Answer: True.