Question

The radical axis of the circles \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] and \[ 2x^2 + 2y^2 + 3x + 8y + 2c = 0 \] touches the circle \[ x^2 + y^2 + 2x + 2y + 1 = 0. \] Then:

• A. \( g = \frac{3}{8} \) \textbf{or} \( f = 1 \)
• B. \( g = \frac{2}{3} \) \textbf{or} \( f = 3 \)
• C. \( g = \frac{1}{2} \) \textbf{or} \( f = 1 \)
• D. \( g = \frac{3}{4} \) \textbf{or} \( f = 2 \)

Hint:

For radical axis problems, subtract the given circle equations and use the touching condition with the third circle to solve for unknowns.

Answer 1

The Correct Option is D

Step 1: Finding the radical axis equation The radical axis is given by subtracting the two circle equations: \[ (x^2 + y^2 + 2gx + 2fy + c) - (2x^2 + 2y^2 + 3x + 8y + 2c) = 0. \] Simplifying, \[ -x^2 - y^2 + (2g - 3)x + (2f - 8)y + c - 2c = 0. \] Step 2: Applying the touching condition For the radical axis to touch the third circle, its perpendicular distance from the center must equal its radius. Solving for \( g \) and \( f \), \[ g = \frac{3}{4}, \quad f = 2. \]

Answer 2

Given two circles:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \] and \[ 2x^2 + 2y^2 + 3x + 8y + 2c = 0 \] Their radical axis touches the circle:
\[ x^2 + y^2 + 2x + 2y + 1 = 0 \] Find the values of \( g \) and \( f \).

Step 1: The radical axis of the two circles is given by subtracting their equations:
\[ (x^2 + y^2 + 2gx + 2fy + c) - \frac{1}{2}(2x^2 + 2y^2 + 3x + 8y + 2c) = 0 \] Multiply second equation by \( \frac{1}{2} \) for consistent comparison:
\[ x^2 + y^2 + 2gx + 2fy + c - (x^2 + y^2 + \frac{3}{2} x + 4 y + c) = 0 \] Simplify:
\[ 2gx + 2fy - \frac{3}{2} x - 4 y = 0 \] \[ (2g - \frac{3}{2}) x + (2f - 4) y = 0 \] Multiply both sides by 2:
\[ (4g - 3) x + (4f - 8) y = 0 \] This is the radical axis.

Step 2: The radical axis touches the circle:
\[ x^2 + y^2 + 2x + 2y + 1 = 0 \] Equation of the line (radical axis) is:
\[ (4g - 3) x + (4f - 8) y = 0 \] or
\[ A x + B y = 0 \] where
\[ A = 4g - 3, \quad B = 4f - 8 \]

Step 3: For a line \( A x + B y + C = 0 \) to be tangent to a circle:
\[ \text{Distance from center} = \text{radius} \] Circle center and radius:
\[ (x + 1)^2 + (y + 1)^2 = 1 \] Center \( C = (-1, -1) \), radius \( r = 1 \).

Step 4: Distance from center to line:
\[ d = \frac{|A(-1) + B(-1) + 0|}{\sqrt{A^2 + B^2}} = \frac{| -A - B |}{\sqrt{A^2 + B^2}} = r = 1 \] Substitute \( A \) and \( B \):
\[ \frac{|-(4g - 3) - (4f - 8)|}{\sqrt{(4g - 3)^2 + (4f - 8)^2}} = 1 \] Simplify numerator:
\[ | -4g + 3 - 4f + 8 | = | 11 - 4(g + f) | \] So:
\[ \frac{|11 - 4(g + f)|}{\sqrt{(4g - 3)^2 + (4f - 8)^2}} = 1 \] Square both sides:
\[ (11 - 4(g + f))^2 = (4g - 3)^2 + (4f - 8)^2 \]

Step 5: Expand RHS:
\[ (4g - 3)^2 + (4f - 8)^2 = 16 g^2 - 24 g + 9 + 16 f^2 - 64 f + 64 = 16 g^2 + 16 f^2 - 24 g - 64 f + 73 \]

Step 6: Expand LHS:
\[ (11 - 4(g + f))^2 = 121 - 88(g + f) + 16 (g + f)^2 = 121 - 88 g - 88 f + 16(g^2 + 2 g f + f^2) \] \[ = 121 - 88 g - 88 f + 16 g^2 + 32 g f + 16 f^2 \]

Step 7: Equate and simplify:
\[ 121 - 88 g - 88 f + 16 g^2 + 32 g f + 16 f^2 = 16 g^2 + 16 f^2 - 24 g - 64 f + 73 \] Subtract RHS from both sides:
\[ 121 - 88 g - 88 f + 16 g^2 + 32 g f + 16 f^2 - 16 g^2 - 16 f^2 + 24 g + 64 f - 73 = 0 \] Simplify:
\[ (121 - 73) + (-88 g + 24 g) + (-88 f + 64 f) + 32 g f = 0 \] \[ 48 - 64 g - 24 f + 32 g f = 0 \]

Step 8: Rearrange:
\[ 32 g f - 64 g - 24 f + 48 = 0 \] Divide entire equation by 8:
\[ 4 g f - 8 g - 3 f + 6 = 0 \] Rearranged:
\[ 4 g f - 8 g - 3 f = -6 \]

Step 9: To find specific values, observe that either \( g = \frac{3}{4} \) or \( f = 2 \) satisfies this equation.

Therefore, the solutions are:
\[ \boxed{ g = \frac{3}{4} \quad \text{or} \quad f = 2 } \]