Question

# If $4{{x}^{2}}+4{{y}^{2}}+4hxy+16x+32y+10=0$ represents a circle, then h is: (a) $5$ (b) $0$ (c) $-2$ (d) $-5$

Hint: Think back to the general form of the equation of a circle. It is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. Now see if there’s any term that contains an xy. Proceed from here.
The general equation of any conic section can be written as :
$a{{x}^{2}}+bxy+c{{y}^{2}}+dx+ey+f=0$ .
Comparing equation ${{C}_{1}}:4{{x}^{2}}+4{{y}^{2}}+4hxy+16x+32y+10=0$, we find :
$a=4,b=4h,c=4,d=16,e=32,f=10$.
For this conic to be the equation of a circle, there are certain conditions that this general equation needs to follow. They are :
1. The value of ${{b}^{2}}-4ac<0$. (This makes sure that the conic is an ellipse).
2. Since a circle is just a special case of an ellipse, having both the lengths of the major and minor axes as equal, another condition for the conic to be a circle is : $a=c$ and $b=0$.

Now, we will check for the conditions to be satisfied one by one.
Condition (1) says that ${{b}^{2}}-4ac<0$.
Substituting for $b,a,c$ in the equation ${{b}^{2}}-4ac<0$, we get :
\begin{align} & {{(4h)}^{2}}-4\times 4\times 4<0 \\ & \Rightarrow 16{{h}^{2}}-64<0 \\ & \Rightarrow 16{{h}^{2}}<64 \\ & \Rightarrow {{h}^{2}}<4 \\ & \Rightarrow {{h}^{2}}-4<0 \\ & \Rightarrow (h-2)(h+2)<0 \\ & \Rightarrow h\in (-2,2) \\ \end{align}
Therefore, $h$ can be any value between $-2$ and $2$ according to the first condition.
However, satisfying the first condition just makes the conic represent an ellipse, rather than a circle.
The next condition to be satisfied is condition (2), which states that for the conic to be a circle, $a=c$ and $b=0$.
Substituting for $a,c,b$, we get :
$4=4$ which is always true.
And, \begin{align} & b=0 \\ & \Rightarrow 4h=0 \\ & \Rightarrow h=0 \\ \end{align}
Thus, for the second condition to be satisfied, the only possible value of $h$ can be $0$.
$h=0$ also satisfies the interval we got from the first condition. It does lie between $-2$ and $2$.
Hence, $h$ should be equal to $0$ for the equation $4{{x}^{2}}+4{{y}^{2}}+4hxy+16x+32y+10=0$ to represent a circle.
Another way of approaching this problem would be to simply compare the equation given to the general form of a circle which is : ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$.
As we can see here, there’s no term containing both $x$ and $y$ in the general equation of a circle. Hence, on comparing we can directly reach the conclusion that the coefficient of $xy=0$ in the equation given. Even then, we would have arrived at the same result.