Question

How do you graph ${{\left( x+1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9$?

Answer 1

Hint: In this question we have an expression which we will convert into the form of the equation of a circle. The general equation of a circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$, where $\left( h,k \right)$ is the center of the circle and $r$ is the radius of the circle. We will simplify the equation to get it in this form, and then compare it with the general form to get the values of $h,k$ and $r$, and then plot then draw the circle on the graph.

Complete step by step solution:
We have the expression given to us as:
$\Rightarrow {{\left( x+1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9$
Now we can write the terms in addition in the form of subtraction of a negative term as:
$\Rightarrow {{\left( x-\left( 1 \right) \right)}^{2}}+{{\left( y-\left( -2 \right) \right)}^{2}}=9$
Now the term $9$ in the right-hand side can be written as the square of the term $3$ as:
$\Rightarrow {{\left( x-\left( 1 \right) \right)}^{2}}+{{\left( y-\left( -2 \right) \right)}^{2}}={{3}^{2}}$
Now we can see that the above expression is in the form of the general equation of a circle which is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
On comparing the terms, we get the values of $h,k$ and $r$ as:
$h=-1$
$k=-2$
$r=3$
Now the center of the circle is the point $\left( h,k \right)$.therefore, we get the radius of the circle as:
$\left( -1,-2 \right)$.
Now on plotting this point on the graph, we get:

Now the radius of the circle is given by $r$, therefore from the expression, we have $r=3$.
Therefore, we will make a circle around point $A$ with the radius $3$ as:

Which is the required solution.

Note: In this question when we took the square root of the term $9$, we only considered the positive value of $3$ and not the negative value since the radius is distance and distance cannot be negative.
The equation of a circle with its center at the origin can be written as ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$, where $r$ is the radius of the circle.