Question

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

Answer 1

Hint: We will first use the method of completing the square and rewrite the given equation. Now, we will compare it to the general equation of a circle and draw it using radius and circle.

Complete step-by-step answer:
We are given that we are required to graph ${x^2} + {y^2} - 6x + 2y - 6 = 0$.
We can write this given equation as:-
$ \Rightarrow {x^2} - 2 \times 3 \times x + {y^2} + 2 \times y - 6 = 0$
Adding and subtracting 9 from the above mentioned equation, we will get the following equation:-
$ \Rightarrow {x^2} - 6x + 9 - 9 + {y^2} + 2 \times y - 6 = 0$ ………………(1)
Now, we know that we have an identity given by: ${(a - b)^2} = {a^2} + {b^2} - 2ab$
Replacing a by x and b by 3, we will then obtain the following equation:-
\[ \Rightarrow {(x - 3)^2} = {x^2} + {3^2} - 2 \times 3 \times x\]
Simplifying the right hand side in the above equation, we will then obtain the following equation:-
\[ \Rightarrow {(x - 3)^2} = {x^2} + 9 - 6x\]
Putting this above mentioned equation in equation number (1), we will then obtain the fact that the given expression can be written as following:-
$ \Rightarrow {(x - 3)^2} - 9 + {y^2} + 2 \times y - 6 = 0$
Adding and subtracting 1 from the above mentioned equation, we will get the following equation:-
$ \Rightarrow {(x - 3)^2} - 9 + {y^2} + 2y + 1 - 1 - 6 = 0$ ………………..(2)
Now, we know that we have an identity given by: ${(a + b)^2} = {a^2} + {b^2} + 2ab$
Replacing a by y and b by 1, we will then obtain the following equation:-
\[ \Rightarrow {(y + 1)^2} = {y^2} + {1^2} + 2 \times 1 \times y\]
Simplifying the right hand side in the above equation, we will then obtain the following equation:-
\[ \Rightarrow {(y + 1)^2} = {y^2} + 1 + 2y\]
Putting this above mentioned equation in equation number (2), we will then obtain the fact that the given expression can be written as following:-
$ \Rightarrow {(x - 3)^2} - 9 + {(y + 1)^2} - 6 - 1 = 0$
Simplifying the terms out of the bracket and taking them to the right hand side, we will then obtain the following expression:-
$ \Rightarrow {(x - 3)^2} + {(y + 1)^2} = 16$
We can also write this equation as:
$ \Rightarrow {(x - 3)^2} + {(y + 1)^2} = {4^2}$
Now, we have a circle with centre (3, - 1) and radius as 4 units.
Drawing this, we will obtain the following:-

Thus, we have required the graph of the given expression.

Note:
The students must note that while finding the radius and centre of the circle, we used the fact that “The general equation of a circle is given by: ${(x - h)^2} + {(y - k)^2} = {r^2}$, where (h, k) is the centre of the circle and r is its radius”.
We compared this to ${(x - 3)^2} + {(y + 1)^2} = {4^2}$ and got the centre as (3, - 1) and the radius as 4 units.