Question

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) $x^{2}+y^{2}-4x-4y-4=0$

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Answer

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.


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