Question

If the circles \[ x^2+y^2-10x+8y+5=0 \] and \[ x^2+y^2+6x-4y+c=0 \] cut each other orthogonally, then the sum of the radii of these circles is

• A. \(14\)
• B. \(10\)
• C. \(12\)
• D. \(9\)

Hint:

Two circles \[ x^2+y^2+2g_1x+2f_1y+c_1=0 \] and \[ x^2+y^2+2g_2x+2f_2y+c_2=0 \] intersect orthogonally if \[ \boxed{2g_1g_2+2f_1f_2=c_1+c_2.} \]

Answer 1

The Correct Option is A

Step 1: Find the radius of the first circle. For \[ x^2+y^2-10x+8y+5=0, \] the centre is \[ (5,-4), \] and \[ r_1=\sqrt{5^2+(-4)^2-5} =\sqrt{36} =6. \]

Step 2: Use the orthogonality condition. For the second circle, \[ x^2+y^2+6x-4y+c=0, \] the centre is \[ (-3,2), \] and \[ r_2^2=9+4-c=13-c. \] If two circles intersect orthogonally, \[ 2g_1g_2+2f_1f_2=c_1+c_2. \] Substituting, \[ 2(-5)(3)+2(4)(-2)=5+c, \] \[ -30-16=5+c, \] \[ c=-51. \] Hence, \[ r_2=\sqrt{13-(-51)} =\sqrt{64} =8. \] Therefore, \[ r_1+r_2=6+8=14. \] Thus, \[ \boxed{14}. \] Hence, the correct option is \(\boxed{(A)}\).