Key Concepts and Formulas

• Equation of a Circle: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Limits: Techniques for evaluating limits, including factorization to resolve indeterminate forms like $\frac{0}{0}$.

Step-by-Step Solution

Step 1: Find the intersection point $(x_c, y_c)$

We are given the lines:

1. $3x + 5y = 1$
2. $(2+c)x + 5c^2y = 1$

Our goal is to solve this system of equations for $x$ and $y$ in terms of $c$.

• Step 1.1: Eliminate $y$ Multiply the first equation by $c^2$ to get $3c^2x + 5c^2y = c^2$. Now subtract this from the second equation: $((2+c)x + 5c^2y) - (3c^2x + 5c^2y) = 1 - c^2$ $(2+c-3c^2)x = 1-c^2$

• Step 1.2: Solve for $x_c$ $x_c = \frac{1-c^2}{2+c-3c^2}$ We factor the numerator and denominator: $x_c = \frac{(1-c)(1+c)}{(1-c)(3c+2)}$ For $c \neq 1$, we can cancel the $(1-c)$ terms: $x_c = \frac{1+c}{3c+2}$

• Step 1.3: Solve for $y_c$ Substitute $x_c$ into the first equation $3x + 5y = 1$: $3\left(\frac{1+c}{3c+2}\right) + 5y_c = 1$ $5y_c = 1 - \frac{3(1+c)}{3c+2}$ $5y_c = \frac{3c+2 - 3 - 3c}{3c+2}$ $5y_c = \frac{-1}{3c+2}$ $y_c = \frac{-1}{5(3c+2)}$

Step 2: Determine the center $(h, k)$ of the circle

We are given that $h = \lim_{c \rightarrow 1} x_c$ and $k = \lim_{c \rightarrow 1} y_c$.

• Step 2.1: Calculate $h$ $h = \lim_{c \rightarrow 1} \frac{1+c}{3c+2} = \frac{1+1}{3(1)+2} = \frac{2}{5}$

• Step 2.2: Calculate $k$ $k = \lim_{c \rightarrow 1} \frac{-1}{5(3c+2)} = \frac{-1}{5(3(1)+2)} = \frac{-1}{5(5)} = -\frac{1}{25}$

Thus, the center of the circle is $(h, k) = \left(\frac{2}{5}, -\frac{1}{25}\right)$.

Step 3: Calculate the radius of the circle

The circle passes through the point $(2, 0)$. The radius $r$ is the distance between the center $(h, k)$ and the point $(2, 0)$.

$r^2 = \left(2 - \frac{2}{5}\right)^2 + \left(0 - \left(-\frac{1}{25}\right)\right)^2$ $r^2 = \left(\frac{10-2}{5}\right)^2 + \left(\frac{1}{25}\right)^2$ $r^2 = \left(\frac{8}{5}\right)^2 + \frac{1}{625}$ $r^2 = \frac{64}{25} + \frac{1}{625} = \frac{64 \cdot 25}{25 \cdot 25} + \frac{1}{625} = \frac{1600}{625} + \frac{1}{625} = \frac{1601}{625}$

Step 4: Find the equation of the circle

The equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$.

$\left(x - \frac{2}{5}\right)^2 + \left(y + \frac{1}{25}\right)^2 = \frac{1601}{625}$ $x^2 - \frac{4}{5}x + \frac{4}{25} + y^2 + \frac{2}{25}y + \frac{1}{625} = \frac{1601}{625}$ Multiply by 625: $625x^2 - 500x + 100 + 625y^2 + 50y + 1 = 1601$ $625x^2 + 625y^2 - 500x + 50y + 101 = 1601$ $625x^2 + 625y^2 - 500x + 50y - 1500 = 0$ Divide by 125: $5x^2 + 5y^2 - 4x + \frac{2}{5}y - 12 = 0$ Multiply by 5: $25x^2 + 25y^2 - 20x + 2y - 60 = 0$

Common Mistakes & Tips

• Be careful when factoring and simplifying expressions, especially when dealing with limits.
• Double-check your calculations to avoid errors, especially when expanding and simplifying the equation of the circle.
• Remember to multiply by a constant to remove fractions and get an equation matching the given options.

Summary

We first found the intersection point of the two lines in terms of $c$, then calculated the limits as $c$ approaches 1 to find the center of the circle. We then used the distance formula to find the radius squared and plugged the center and radius into the equation of a circle. Finally, we multiplied by a constant to match the form of the answer choices. The equation of the circle is $5x^2+5y^2-4x+\frac{2}{5}y-12 = 0$ which can also be written as $25x^2+25y^2-20x+2y-60 = 0$.

Final Answer

The final answer is \boxed{25 x^2+25 y^2-20 x+2 y-60=0}, which corresponds to option (B).