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# 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.

Note: The general equation listed in this sum is an equation that is true for all conic sections. Imposing restrictions on this general form gives us different types of conics, like circles, ellipses, parabolas, etc. Just like we solved for the equation to be a circle right now, it can also be made to represent a parabola or any other conic simply by tweaking the variables.
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.
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CBSE Class 10 Maths Chapter 10 - Circles Formula  Circle    Circumference of a Circle  Tangent of a Circle  Tangent to a Circle  Equation of A Circle  Secant of a Circle  Sector of a Circle  Area of a Circle  