Question

How do you determine circle, parabola, ellipse or hyperbola from the equation ${{x}^{2}}+{{y}^{2}}-16x+18y-11=0$?

Answer 1

Hint: We will first try to group the terms present in the equation and then complete the square for both x and y terms. Then we will compare the obtained equation with the general equations of circle, parabola, ellipse and hyperbola to get the desired answer.

Complete step-by-step solution:
We have been given an equation ${{x}^{2}}+{{y}^{2}}-16x+18y-11=0$.
We have to determine whether the given equation is of circle, parabola, ellipse or hyperbola.
Now, let us group the x and y terms then we will get
$\Rightarrow {{x}^{2}}-16x+{{y}^{2}}+18y=11$
Now, to complete the square for x we need to add 64 both sides of the equation. Then we will get
$\begin{align}
  & \Rightarrow {{x}^{2}}-16x+64+{{y}^{2}}+18y=11+64 \\
 & \Rightarrow {{x}^{2}}-8\times 2x+{{8}^{2}}+{{y}^{2}}+18y=11+64 \\
 & \Rightarrow {{\left( x-8 \right)}^{2}}+{{y}^{2}}+18y=75 \\
\end{align}$
Now, to complete the square for y we need to add 81 both sides of the equation. Then we will get
$\begin{align}
  & \Rightarrow {{\left( x-8 \right)}^{2}}+{{y}^{2}}+18y+81=75+81 \\
 & \Rightarrow {{\left( x-8 \right)}^{2}}+{{y}^{2}}+2\times 9y+{{9}^{2}}=156 \\
 & \Rightarrow {{\left( x-8 \right)}^{2}}+{{\left( y+9 \right)}^{2}}=156 \\
\end{align}$
Now, we know that the above obtained equation is similar to the standard equation of circle which is given as ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$

Hence the given equation represents the circle.


Note: Alternatively by analyzing some points we can determine whether the equation belongs to a circle, parabola, ellipse or hyperbola.
If the coefficients of both ${{x}^{2}}$ and ${{y}^{2}}$ are the same then it is a circle. If the coefficients of both ${{x}^{2}}$ and ${{y}^{2}}$ are different and have positive or negative value then it is an ellipse. If there is only one squared term then it is a parabola and if there is one squared term and has negative coefficient then it is a hyperbola.
From the above points the given equation has the same coefficients of both ${{x}^{2}}$ and ${{y}^{2}}$ so it is an equation of circle.