Question

Let a circle passing through (2, 0) have its centre at the point \( (h, k) \). Let \( (x_c, y_c) \) be the point of intersection of the lines \( 3x + 5y = 1 \) and \( (2 + c)x + 5c^2y = 1 \). If \( h = \lim_{c \to 1} x_c \) and \( k = \lim_{c \to 1} y_c \), then the equation of the circle is:

• A. \( 25x^2 + 25y^2 - 20x + 2y - 60 = 0 \)
• B. \( 5x^2 + 5y^2 - 4x - 2y - 12 = 0 \)
• C. \( 25x^2 + 25y^2 - 2x + 2y - 60 = 0 \)
• D. \( 5x^2 + 5y^2 - 4x + 2y - 12 = 0 \)

Answer 1

The Correct Option is A

Let the circle pass through the point \( (2, 0) \) with its center at \( (h, k) \). The point of intersection of the lines \[ 3x + 5y = 1 \quad \text{and} \quad (2 + c)x + 5c^2y = 1 \] is given as \( (x_c, y_c) \), where: \[ x = \frac{1 - c^2}{2 + c - 3c^2}, \quad y = \frac{1 - 3c}{5(2 + c - 3c^2)}. \]

Step 1: Limits of \( h \) and \( k \)

\[ h = \lim_{c \to 1} x = \lim_{c \to 1} \frac{(1 - c)(1 + c)}{(1 - c)(2 + 3c)} = \frac{2}{5}, \] 
\[ k = \lim_{c \to 1} y = \lim_{c \to 1} \frac{c - 1}{-5(c - 1)(3c + 2)} = -\frac{1}{25}. \] 
Thus, the center of the circle is: \[ \left( \frac{2}{5}, -\frac{1}{25} \right). \]

Step 2: Radius of the circle

Using the distance formula, the radius is: 
\[ r = \sqrt{\left( 2 - \frac{2}{5} \right)^2 + \left( 0 - \left(-\frac{1}{25}\right)\right)^2}. \] 
Simplify: 
\[ r = \sqrt{\left(\frac{10}{5} - \frac{2}{5}\right)^2 + \left(\frac{1}{25}\right)^2} = \sqrt{\left(\frac{8}{5}\right)^2 + \left(\frac{1}{25}\right)^2}. \]
\[ r = \sqrt{\frac{64}{25} + \frac{1}{625}} = \sqrt{\frac{1600 + 1}{625}} = \sqrt{\frac{1601}{625}} = \frac{\sqrt{1601}}{25}. \]

Step 3: Equation of the circle

The general equation of the circle is:
\[ \left(x - \frac{2}{5}\right)^2 + \left(y + \frac{1}{25}\right)^2 = \left(\frac{\sqrt{1601}}{25}\right)^2. \]
Simplify:
\[ \left(x - \frac{2}{5}\right)^2 + \left(y + \frac{1}{25}\right)^2 = \frac{1601}{625}. \]
Multiply through by 625:
\[ 25\left(x - \frac{2}{5}\right)^2 + 25\left(y + \frac{1}{25}\right)^2 = 1601. \]
Expand:
\[ 25x^2 - 20x + 4 + 25y^2 + 2y + \frac{1}{25} = 1601. \]
Simplify:
\[ 25x^2 + 25y^2 - 20x + 2y - 60 = 0. \]

Final Answer:

\[ \boxed{25x^2 + 25y^2 - 20x + 2y - 60 = 0.} \]