Question

The radius of the circle passing through the points of intersection of the circles \( x^2+y^2+2x+4y+1=0 \), \( x^2+y^2-2x-4y-4=0 \), and intersecting the circle \( x^2+y^2=6 \) orthogonally is:

• A. \( \sqrt{19} \)
• B. \( 5 \)
• C. \( \sqrt{39} \)
• D. \( 4 \)

Hint:

For finding the equation of a circle that intersects another circle orthogonally, use the condition: \[ 2g g' + 2f f' = c + c' \] where \( g, f, c \) are the coefficients of the required circle, and \( g', f', c' \) are those of the given circle.

Answer 1

The Correct Option is A

The two given circles: \[ x^2 + y^2 + 2x + 4y + 1 = 0 \] \[ x^2 + y^2 - 2x - 4y - 4 = 0 \] Subtracting to find their radical axis: \[ (2x + 4y + 1) - (-2x - 4y - 4) = 0 \] \[ 4x + 8y + 5 = 0 \] The required circle intersects \( x^2 + y^2 = 6 \) orthogonally, meaning: \[ 2g \times 0 + 2f \times 0 - c = -6 \] Solving for \( r \): \[ r = \sqrt{19} \] Thus, the correct answer is \( \sqrt{19} \).