Hello!
The answer is:
The center of the circle is located on the point (-9,-2) and the radius is 6 units.
Why?
To solve the problem, we need to use the given equation which is in the general form.
We are given the circle:
[tex]x^{2}+y^{2}+18x+4y+49=0[/tex]
We know that a circle can be written in the following form:
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
Where,
h is the x-coordinate of the center of the circle
k is the y-coordinate of the center of the circle
r is the radius of the circle.
So, to find the center and the radius, we need to perform the following steps:
- Moving the constant to the other side of the equation:
[tex]x^{2}+y^{2}+18x+4y=-49[/tex]
- Grouping the terms:
[tex]x^{2}+18x+y^{2}+4y=-49[/tex]
- Completing squares for both variables, we have:
We need to sum to each side of the equation the following term:
[tex](\frac{b}{2})^{2}[/tex]
Where, b, for this case, will the coefficients for both terms that have linear variables (x and y)
So, the variable "x", we have:
[tex]x^{2} +18x[/tex]
Where,
[tex]b=18[/tex]
Then,
[tex](\frac{18}{2})^{2}=(9)^{2}=81[/tex]
So, we need to add the number 81 to each side of the circle equation.
Now, for the variable "y", we have:
[tex]y^{2} +4y[/tex]
Where,
[tex]b=4[/tex]
[tex](\frac{4}{2})^{2}=(2)^{2}=4[/tex]
So, we need to add the number 4 to each side of the circle equation.
Therefore, we have:
[tex](x^{2}+18x+81)+(y^{2}+4y+4)=-49+81+4[/tex]
Then, factoring, we have that the expression will be:
[tex](x+9)^{2}+(y+2)^{2}=36[/tex]
- Writing the standard form of the circle:
Now, from the simplified expression (after factoring), we can find the equation of the circle in the standard form:
[tex](x+9)^{2}+(y+2)^{2}=36[/tex]
Is also equal to:
[tex](x-(-9))^{2}+(y-(-2))^{2}=36[/tex]
Where,
[tex]h=-9\\k=-2\\r=\sqrt{36}=6[/tex]
Hence, the center of the circle is located on the point (-9,-2) and the radius is 6 units.
Have a nice day!
