Question

The radical axis of the circles \[ x^2+y^2+4x+6y+7=0 \] and \[ 4x^2+4y^2+8x+12y-24=0 \] is a tangent to the circle \[ x^2+y^2=13 \] at a point \((\alpha,\beta)\), then \(\alpha+\beta=\)

• A. \(0\)
• B. \(-\dfrac52\)
• C. \(1\)
• D. \(-5\)

Hint:

The point of contact of a tangent to the circle \[ x^2+y^2=r^2 \] lies on the radius perpendicular to the tangent line. The radius vector is parallel to the normal vector of the tangent.

Answer 1

The Correct Option is D

Step 1: Find the radical axis. Divide the second circle by \(4\): \[ x^2+y^2+2x+3y-6=0. \] Subtracting this from the first circle, \[ (4x-2x)+(6y-3y)+7+6=0, \] \[ 2x+3y+13=0. \] Hence, the radical axis is \[ 2x+3y+13=0. \]

Step 2: Use the tangent property. The circle \[ x^2+y^2=13 \] has centre \[ (0,0). \] Since \[ 2x+3y+13=0 \] is tangent to the circle, the point of contact lies on the radius perpendicular to the tangent. The normal vector of the tangent is \[ (2,3). \] Hence, \[ (\alpha,\beta) = -\frac{\sqrt{13}}{\sqrt{2^2+3^2}}(2,3) = (-2,-3). \]

Step 3: Find the required sum. Therefore, \[ \alpha+\beta = -2+(-3) = -5. \] Hence, \[ \boxed{-5}. \] Thus, \[ \boxed{(D)} \] is the correct answer.