Hello!
The answer is:
The center of the circle is the point (-9,-2) and the circle has a radius of 6 units.
Why?
To solve the problem, we need to use the given equation which is in the general form, and then, use it transform it to the standard form in order to find the center of the circle and its radius.
So,
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 (x and y):
[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 write 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 the point (-9,-2) and the circle has a radius of 6 units.
Have a nice day!
