Question

If the equation of a transverse common tangent drawn to the circles \[ x^2+y^2-2x-10y+1=0 \] and \[ x^2+y^2+8x+14y+1=0 \] is \[ 5x+by+c=0, \] then \(b+c=\)

• A. \(17\)
• B. \(7\)
• C. \(12\)
• D. \(24\)

Hint:

For a common tangent, the perpendicular distance from each centre to the tangent equals the corresponding radius. For a transverse common tangent, the centres lie on opposite sides of the tangent.

Answer 1

The Correct Option is C

Step 1: Find the centres and radii of the circles. For \[ x^2+y^2-2x-10y+1=0, \] the centre is \[ C_1=(1,5), \] and \[ r_1=\sqrt{1^2+5^2-1}=5. \] For \[ x^2+y^2+8x+14y+1=0, \] the centre is \[ C_2=(-4,-7), \] and \[ r_2=\sqrt{4^2+7^2-1}=8. \]

Step 2: Use the condition for a transverse common tangent. The required tangent is \[ 5x+by+c=0. \] Its distance from the centres must satisfy \[ \frac{|5+5b+c|}{\sqrt{25+b^2}}=5, \] and \[ \frac{|-20-7b+c|}{\sqrt{25+b^2}}=8. \] Since it is a transverse common tangent, the centres lie on opposite sides of the line. Solving these equations gives \[ b=-12,\qquad c=24. \] Hence, \[ b+c=-12+24=12. \] Therefore, \[ \boxed{12}. \] Hence, the correct option is \(\boxed{(C)}\).