The equation of a circle which touches both axes and the line 3x-4y+8=0 and whose center lies in the third quadrant is A) x^2+y^2-4x+4y-4=0 B) x^2+y^2-4x+4y+4=0 C) x^2+y^2+4x+4y+4=0 D) x^2+y^2-4x-4y-4=0

To solve this problem, we need to find the equation of a circle that touches both the x and y axes, as well as the line $3x-4y+8=0$. Additionally, the center of the circle is located in the third quadrant.

Steps:

1. Equation Formulation: The general form of a circle that touches both axes and has its center in the third quadrant can be written as: $$(x + a)^2 + (y + a)^2 = a^2$$ Here, $a$ is the radius of the circle, and $-a$ for both x and y coordinates of the center indicate that the center is in the third quadrant. Expanding this: $$x^2 + y^2 + 2ax + 2ay + 2a^2 = a^2$$ This simplifies to: $$x^2 + y^2 + 2ax + 2ay + a^2 = 0$$

2. Line Intersection Condition: This circle must also be tangent to the line $3x - 4y + 8 = 0$. The distance from the center of the circle $(-a, -a)$ to this line must be equal to the radius $a$ of the circle. Using the distance formula between a point and a line: $$d = \left| \frac{3(-a) - 4(-a) + 8}{\sqrt{9 + 16}} \right| = a$$ Simplifying, we get: $$\left| \frac{8 - a}{5} \right| = a$$ Solving for $a$, given that the center is in the third quadrant (both $a$ and coordinate values are negative), $a = 2$.

3. Substituting $a$: Plugging $a = 2$ into our circle equation: $$x^2 + y^2 + 2(-2)x + 2(-2)y + 2^2 = 0$$ leads to: $$x^2 + y^2 + 4x + 4y + 4 = 0$$

Conclusion:

The equation for the circle is: $$\mathbf{x^2 + y^2 + 4x + 4y + 4 = 0}$$ Thus, the correct answer is Option C.