Question

How do you write \[{x^2} + {y^2} + 8x + 7 = 0\] in standard form?

Answer 1

Hint: Here, we will group the terms of the equation and compare the equation with the general form of a quadratic equation to get the respective coefficient of the variables and constant term. Then by completing the square method and simplifying the equation by using the algebraic identity, we will find the standard form of the given equation.

Formula Used:
We will use the following formula:
1. Completing the square method is given by the formula \[a{x^2} + bx + c = {\left( {x \pm \dfrac{b}{2}} \right)^2} + c + {\left( {\dfrac{b}{2}} \right)^2}\]
2. The square of the sum of the numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
3. Standard form of the circle \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}\] with the center \[\left( {h,k} \right)\] and radius \[a\]

Complete step by step solution:
We are given an equation \[{x^2} + {y^2} + 8x + 7 = 0\].
Now, we will group the \[x\]- term, \[y\]- term on one side of the equation and the constant term on the other side of the equation. Therefore, we get
\[ \Rightarrow \left( {{x^2} + 8x} \right) + {y^2} + 7 = 0\]
Completing the square method is given by the formula \[a{x^2} + bx + c = {\left( {x \pm \dfrac{b}{2}} \right)^2} + c + {\left( {\dfrac{b}{2}} \right)^2}\]
Quadratic equation \[\left( {{x^2} + 8x + 7} \right)\] is of the form \[a{x^2} + bx + c = 0\].
Comparing with the given quadratic equation, we get
\[\begin{array}{l}a = 1\\b = 8\\c = - 7\end{array}\]
Now, by using the completing the square method for the \[x\]- term, we get
\[\left( {{x^2} + \left( {\dfrac{8}{2}} \right)x + {{\left( {\dfrac{8}{2}} \right)}^2}} \right) + {y^2} = - 7 + {\left( {\dfrac{8}{2}} \right)^2}\]
Now, by simplifying the equation, we get
\[ \Rightarrow \left( {{x^2} + 4x + {{\left( 4 \right)}^2}} \right) + {y^2} = - 7 + {\left( 4 \right)^2}\]
Applying the exponent on the terms, we get
\[ \Rightarrow \left( {{x^2} + 4x + 16} \right) + {y^2} = - 7 + 16\]
Subtracting the like terms on RHS, we get
\[ \Rightarrow \left( {{x^2} + 4x + 16} \right) + {y^2} = 9\]
The square of the sum of the numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Now, by using the algebraic identity, we get
\[ \Rightarrow {\left( {x + 4} \right)^2} + {y^2} = 9\]
\[ \Rightarrow {\left( {x + 4} \right)^2} + {y^2} = {3^2}\]
The above equation is a standard equation of a circle the center \[\left( { - 4,0} \right)\] and radius 3.
This is the standard form of the circle \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}\] with the center \[\left( {h,k} \right)\] and radius \[a\]

Therefore, the standard form of \[{x^2} + {y^2} + 8x + 7 = 0\] is \[{\left( {x + 4} \right)^2} + {y^2} = 9\].

Note:
We should remember that when performing the completing the square method, there will always be a constant which is the square of the half of the coefficient of \[x\]-term will be added to both the sides of the equation, so that the equation is balanced. Standard form of the equation of the circle is a way of expressing the circle in the coordinate plane. Whenever we are converting the equation to standard form, we can always use the completing the square method in terms of \[x\] and \[y\] separately.