Question

If the circles \[ x^2+y^2+2gx+6y+4=0 \] and \[ x^2+y^2-gx-2y-14=0 \] cut each other orthogonally for a positive integral value of \(g\), then the radical axis of the two circles is:

• A. \(3x+8y+18=0\)
• B. \(9x+8y+18=0\)
• C. \(3x+4y+9=0\)
• D. \(6x+4y+9=0\)

Hint:

The radical axis of two circles is obtained immediately by subtracting their equations. First determine any unknown parameters, then subtract.

Answer 1

The Correct Option is C

Concept: Two circles intersect orthogonally if \[ 2g_1g_2+2f_1f_2=c_1+c_2. \] After finding the value of the parameter \(g\), we obtain the radical axis simply by subtracting the two circle equations.

Step 1: Identify coefficients. For \[ x^2+y^2+2gx+6y+4=0, \] we have \[ g_1=g,\qquad f_1=3,\qquad c_1=4. \] For \[ x^2+y^2-gx-2y-14=0, \] \[ 2g_2=-g \] \[ g_2=-\frac g2, \] and \[ f_2=-1,\qquad c_2=-14. \]

Step 2: Use orthogonality condition. Applying \[ 2g_1g_2+2f_1f_2=c_1+c_2, \] gives \[ 2\left(g\right)\left(-\frac g2\right) + 2(3)(-1) = 4+(-14). \] \[ -g^2-6=-10. \] \[ g^2=4. \] Since \(g\) is positive, \[ g=2. \]

Step 3: Substitute \(g=2\). The circles become \[ x^2+y^2+4x+6y+4=0 \] and \[ x^2+y^2-2x-2y-14=0. \]

Step 4: Obtain the radical axis. Subtracting, \[ (4x+6y+4)-(-2x-2y-14)=0. \] \[ 6x+8y+18=0. \] Dividing by \(2\), \[ 3x+4y+9=0. \] \[ \boxed{3x+4y+9=0}. \]