Find the center (h,k) and radius r of the circle and then use these to (a) graph the circle and (b) find the intercepts, if any.

3x^2+36x+3y^2=0

3x^2+36x+3y^2=0

x^2+12x+y^2=0
(x^2+12x+36) + y^2 = 36
(x+6)^2 + y^2 = 36

hat should make it easy to get the required info.

To find the center (h, k) and radius r of the circle represented by the equation 3x^2 + 36x + 3y^2 = 0, we can rewrite the equation in standard form: x^2 + y^2 + Dx + Ey + F = 0, where D and E represent the coefficients of x and y respectively.

We start by dividing the entire equation by the coefficient of y^2, which is 3. This gives us:

x^2/3 + 12x + y^2/3 = 0

Next, we rearrange the terms and complete the squares for both x and y. For x, we complete the square by adding half the coefficient of x, squared. For y, we do the same.

x^2/3 + 12x + 36 + y^2/3 = 36
(x^2/3 + 12x + 36) + (y^2/3) = 36

Now, we have created perfect squares for both x and y terms:

(x + 6)^2 + (y^2/3) = 36

From this, we can see that the center of the circle is given by the values inside the parentheses, which are (-6, 0). The radius of the circle is given by the square root of the constant term on the right side of the equation, which is 6.

Therefore, the center of the circle is (-6, 0) and the radius is 6.

To graph the circle, plot a point at the center (-6, 0) and sketch a circle with a radius of 6 units around this point.

To find the intercepts of the circle, we can substitute y = 0 into the equation to find the x-intercepts and substitute x = 0 to find the y-intercepts.

When y = 0:
x^2/3 + 12x + y^2/3 = 36
x^2/3 + 12x = 36
x^2 + 36x = 108
x^2 + 36x - 108 = 0

We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. From there, we can determine the x-intercepts.

Similarly, we can substitute x = 0 to find the y-intercepts.

By performing these calculations, we can find any intercepts the circle may have.