Question

Need this badly!!!!!!

Answer 1

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:

$x^{2}+y^{2}+18x+4y+49=0$

We know that a circle can be written in the following form:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

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:

$x^{2}+y^{2}+18x+4y=-49$

- Grouping the terms:

$x^{2}+18x+y^{2}+4y=-49$

- Completing squares for both variables, we have:

We need to sum to each side of the equation the following term:

$(\frac{b}{2})^{2}$

Where, b, for this case, will the coefficients for both terms that have linear variables (x and y)

So, the variable "x", we have:

$x^{2} +18x$

Where,

$b=18$

Then,

$(\frac{18}{2})^{2}=(9)^{2}=81$

So, we need to add the number 81 to each side of the circle equation.

Now, for the variable "y", we have:

$y^{2} +4y$

Where,

$b=4$

$(\frac{4}{2})^{2}=(2)^{2}=4$

So, we need to add the number 4 to each side of the circle equation.

Therefore, we have:

$(x^{2}+18x+81)+(y^{2}+4y+4)=-49+81+4$

Then, factoring, we have that the expression will be:

$(x+9)^{2}+(y+2)^{2}=36$

- 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:

$(x+9)^{2}+(y+2)^{2}=36$

Is also equal to:

$(x-(-9))^{2}+(y-(-2))^{2}=36$

Where,

$h=-9\\k=-2\\r=\sqrt{36}=6$

Hence, the center of the circle is located on the point (-9,-2) and the radius is 6 units.

Have a nice day!