Let a and b non-zero real numbers. Then the Equation $\left(a{x}^2+b{y}^2+c\right)\left(x^2-5xy+6y^2\right)=0$ represents.

$$\begin{gathered} \text{ When }a\text{ and }b\text{ are of the same sign and }c=0\text{, given equation reduce to} \\ \left(a{x}^2+b{y}^2\right)\left(x^2-5xy+6y^2\right)=0 \\ \Rightarrow x^2-5xy+6y^2=0\left[\begin{array}{r}\because x^2+b{y}^2\ne 0\text{ when }a\text{ and }b\\ \text{ are of the same sign }\end{array}\right] \\ \Rightarrow (x-2y)(x-3y)=0\Rightarrow x=2y,x=3y \\ \text{ So, the given equation represents a pair of straight lines. When } \\ a=b\text{ and the sign of }c\text{ is opposite to that of }a\text{. The given equation } \\ reduces to \\ \left(a{x}^2+a{y}^2+c\right)\left(x^2-5xy+6y^2\right)=0 \\ \Rightarrow a{x}^2+a{y}^2+c=0\text{ or }{x}^2-5xy+6y^2=0 \\ \Rightarrow x^2+y^2=-\frac{c}{a},x=2y,x=3y \\ \text{ So, the given equation represents a circle of radius }\sqrt{\frac{-c}{a}}\text{ and a pair of straight lines. } \end{gathered}$$