Question

How do you find the centre and radius of the circle ${x^2} + {y^2} + 4x - 4y - 1 = 0$?

Answer 1

Hint: First we have to move $1$ to the right side of the given equation. Next, we have to regroup the terms in such a way that terms containing $x$ and $y$ are in separate parentheses. Next, we have to create a trinomial square on the left side of the equation. For this we have to find a value that is equal to the square of half of $4$. Next, we have to add the term to both parenthesis of the given equation and the right side of the equation. Next, we have to factor the trinomial using algebraic identities. Finally, we have to compare the obtained equation of the circle to the standard equation of the circle and find the value of the numbers $h$, $k$ and $a$ , then we will get the center and radius of the given circle.

Formula used:
Standard Equation of a Circle: The equation of a circle with center at $\left( {h,k} \right)$ and radius equal to $a$, is
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}$……(i)
Algebraic Identity:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$……(ii)
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$……(iii)
Where, $a$ and $b$ are any two numbers.

Complete step by step solution:
First, we have to move $1$to the right side of the given equation. Thus, adding $1$ to both sides of the equation.
${x^2} + {y^2} + 4x - 4y = 1$
Now, we have to regroup the terms in such a way that terms containing $x$ and $y$ are in separate parentheses.
$\left( {{x^2} + 4x} \right) + \left( {{y^2} - 4y} \right) = 1$
Now, we have to create a trinomial square on the left side of the equation. For this we have to find a value that is equal to the square of half of $4$.
${\left( {\dfrac{4}{2}} \right)^2} = {\left( 2 \right)^2}$
Now, we have to add the term to both parenthesis of the given equation and the right side of the equation.
$\left( {{x^2} + 4x + {{\left( 2 \right)}^2}} \right) + \left( {{y^2} - 4y + {{\left( 2 \right)}^2}} \right) = 1 + {\left( 2 \right)^2} + {\left( 2 \right)^2}$
It can be written as
\[ \Rightarrow \left( {{x^2} + 4x + {{\left( 2 \right)}^2}} \right) + \left( {{y^2} - 4y + {{\left( 2 \right)}^2}} \right) = 1 + 4 + 4\]
\[ \Rightarrow \left( {{x^2} + 4x + {{\left( 2 \right)}^2}} \right) + \left( {{y^2} - 4y + {{\left( 2 \right)}^2}} \right) = 9\]
\[ \Rightarrow \left( {{x^2} + 2 \cdot 2 \cdot x + {{\left( 2 \right)}^2}} \right) + \left( {{y^2} - 2 \cdot 2 \cdot x + {{\left( 2 \right)}^2}} \right) = {\left( 3 \right)^2}\]
Now, we have to factor the perfect trinomial into ${\left( {x + 2} \right)^2}$ and ${\left( {y - 2} \right)^2}$ using (ii) and (iii) algebraic identities respectively.
${\left( {x + 2} \right)^2} + {\left( {y - 2} \right)^2} = {\left( 3 \right)^2}$
Now we have to compare the above equation of the circle to the standard equation of the circle and find the value of the numbers $h$, $k$ and $a$.
Comparing ${\left( {x + 2} \right)^2} + {\left( {y - 2} \right)^2} = {\left( 3 \right)^2}$ with ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}$, we get
$h = - 2$, $k = 2$ and $a = 3$
Since, $\left( {h,k} \right)$ is the center of the circle, $a$ is the radius of the circle.
Final solution: Therefore, $\left( { - 2,2} \right)$ is the centre and $3$ is the radius of the given circle.

Note:
We can directly find the radius and centre of a circle by comparing it with the General Equation of a Circle.
General Equation of Circle:
The equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$ always represents a circle whose centre is $\left( { - g, - f} \right)$ and radius is $\sqrt {{g^2} + {f^2} - c} $……(iv)
Step by step solution:
First, we have to compare ${x^2} + {y^2} + 4x - 4y - 1 = 0$ with ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and find the value of $g,f,c$.
$g = 2,f = - 2,c = - 1$
Now we have to find the centre and radius of a given circle by putting the value of $g,f,c$ in (iv).
Since, $\left( { - g, - f} \right)$ is the centre and $\sqrt {{g^2} + {f^2} - c} $ is the radius of the circle.
Here, $g = 2,f = - 2,c = - 1$.
So, $\left( { - g, - f} \right) = \left( { - 2,2} \right)$
$\sqrt {{g^2} + {f^2} - c} = \sqrt {{2^2} + {{\left( { - 2} \right)}^2} - \left( { - 1} \right)} $
$ \Rightarrow \sqrt {{g^2} + {f^2} - c} = \sqrt {4 + 4 + 1} $
$ \Rightarrow \sqrt {{g^2} + {f^2} - c} = \sqrt 9 $
$ \Rightarrow \sqrt {{g^2} + {f^2} - c} = 3$
We have taken only positive roots as radius can’t be negative.
Final solution: Therefore, $\left( { - 2,2} \right)$ is the centre and $3$ is the radius of the given circle.