Question

How do you graph ${x^2} + {y^2} + x - 6y + 9 = 0$?

Answer 1

Hint: According to the question we have to graph the expression ${x^2} + {y^2} + x - 6y + 9 = 0$ which is as mentioned in the question. So, first of all to determine the graph we have to make the given expression in the form of the whole square both of x and y.
Now, as we can see that the given expression is the form of the equation of the circle as ${x^2} + {y^2} + 2ax + 2by + c = 0$. Hence, to obtain the expression in the form of the whole square we have to add $\dfrac{1}{4}$ and subtract $\dfrac{1}{4}$ to make the expression in the whole square of x.

Formula used: $ {a^2} + {b^2} + 2ab = {(a + b)^2}.............(B)$
Now, to make the expression in the form of the whole square of y we have to make 9 in the form of the whole square in the expression.
$ \Rightarrow {a^2} + {b^2} - 2ab = {(a - b)^2}.............(C)$
Now, we have to take all the whole square of x and y in any one side of the expression and all the integers in the other side of the expression.
Now, to obtain the radius and the centre of the circle we have to compare the obtained equation of the circle with the general form of the expression which is as mentioned below:
$ \Rightarrow {(x - a)^2} + {(y - b)^2} = {r^2}...................(D)$
Where, r is the centre of the circle and (a, b) is the radius of the circle.

Complete step-by-step solution:
Step 1: First of all to determine the graph we have to make the given expression in the form of the whole square both of x and y which is as mentioned in the solution hint.
Step 2: Now, as we can see that the given expression is the form of the equation of the circle as ${x^2} + {y^2} + 2ax + 2by + c = 0$. Hence, to obtain the expression in the form of the whole square we have to add $\dfrac{1}{4}$ and subtract $\dfrac{1}{4}$ to make the expression in the whole square of x. Hence,
\[ \Rightarrow {x^2} + x + \dfrac{1}{4} + {y^2} - 6y + 9 - \dfrac{1}{4} = 0\]
Step 3: Now, to make the expression as obtained in the solution step 2, whole square in form of x we have to use the formula (B) which is as mentioned in the question. Hence,
\[ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {y^2} - 6y + 9 - \dfrac{1}{4} = 0\]
Step 4: Now, to make the expression in the form of the whole square of y we have to make 9 in the form of the whole square in the expression. Hence,
\[ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {y^2} - 6y + {3^2} - \dfrac{1}{4} = 0\]
Step 5: Now, we have to use the formula (C) to determine the whole square of y which is as mentioned in the solution hint. Hence,
\[ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {(y - 3)^2} - \dfrac{1}{4} = 0\]
Step 6: Now, we have to take all the whole square of x and y in any one side of the expression and all the integers in the other side of the expression. Hence,
\[ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {(y - 3)^2} = \dfrac{1}{4}\]
Now, to obtain the radius and the centre of the circle we have to compare the obtained equation of the circle with the general form of the expression (D) which is as mentioned in the solution hint.
$ \Rightarrow $Centre$ = \left( {\dfrac{1}{2}, - 3} \right)$
$ \Rightarrow $Radius\[ = \dfrac{1}{2}\]

Hence, with the help of the formula (A) and (B) we have determined the required graph for the given expression which is ${x^2} + {y^2} + x - 6y + 9 = 0$ as below:

Note: To determine the points to plot in the graph it is necessary that we have to make the expression in the form of the whole square of x and y which can be done by adding and subtracting $\dfrac{1}{4}$ in the expression.
To determine the centre and radius of the circle it is necessary that we have to compare the expression obtained with the general form of the equation of the circle.