Question:
The sum of the slopes of the common tangents drawn to the circles \( x^{2}+y^{2}+4x-2y-11=0 \) and \( x^{2}+y^{2}-2x+6y+6=0 \) is
The sum of the slopes of the common tangents drawn to the circles \( x^{2}+y^{2}+4x-2y-11=0 \) and \( x^{2}+y^{2}-2x+6y+6=0 \) is
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When asked for the sum or product of slopes of common tangents, you don't need to waste time calculating the exact values of the individual slopes. Just apply Viet's formulas directly to the resulting quadratic slope equation.
Updated On: Jun 8, 2026
- \( \frac{24}{5} \)
- \( -\frac{24}{5} \)
- \( \frac{8}{3} \)
- \( -\frac{8}{3} \)
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The Correct Option is B
Solution and Explanation
Concept:
Common tangents of two circles pass through their centers of similitude. The internal and external centers of similitude divide the line segment joining the centers of the two circles in the ratio of their radii.
Step 1: Extracting properties of both circles.
• Circle 1: Center \( C_1(-2, 1) \), Radius \( r_1 = \sqrt{(-2)^2 + 1^2 - (-11)} = \sqrt{4+1+11} = 4 \)
• Circle 2: Center \( C_2(1, -3) \), Radius \( r_2 = \sqrt{1^2 + (-3)^2 - 6} = \sqrt{1+9-6} = 2 \)
Step 2: Finding the external center of similitude.
The external center of similitude \( T \) divides \( C_1C_2 \) externally in the ratio \( r_1 : r_2 = 4 : 2 = 2 : 1 \): \[ T = \left( \frac{2(1) - 1(-2)}{2 - 1}, \frac{2(-3) - 1(1)}{2 - 1} \right) = (2 + 2, -6 - 1) = (4, -7) \]
Step 3: Formulating the slope equation.
Any line through \( (4, -7) \) with slope \( m \) is \( y + 7 = m(x - 4) \implies mx - y - (4m + 7) = 0 \). Setting its distance from \( C_2(1, -3) \) equal to its radius \( r_2 = 2 \): \[ \frac{|m(1) - (-3) - 4m - 7|}{\sqrt{m^2 + 1}} = 2 \implies \frac{|-3m - 4|}{\sqrt{m^2 + 1}} = 2 \] Squaring both sides: \[ (3m + 4)^2 = 4(m^2 + 1) \implies 9m^2 + 24m + 16 = 4m^2 + 4 \implies 5m^2 + 24m + 12 = 0 \] The sum of the slopes (roots of this quadratic equation) is given by \( -\frac{\text{Coeff of } m}{\text{Coeff of } m^2} \): \[ m_1 + m_2 = -\frac{24}{5} \] This precisely matches Option (B).
Step 1: Extracting properties of both circles.
• Circle 1: Center \( C_1(-2, 1) \), Radius \( r_1 = \sqrt{(-2)^2 + 1^2 - (-11)} = \sqrt{4+1+11} = 4 \)
• Circle 2: Center \( C_2(1, -3) \), Radius \( r_2 = \sqrt{1^2 + (-3)^2 - 6} = \sqrt{1+9-6} = 2 \)
Step 2: Finding the external center of similitude.
The external center of similitude \( T \) divides \( C_1C_2 \) externally in the ratio \( r_1 : r_2 = 4 : 2 = 2 : 1 \): \[ T = \left( \frac{2(1) - 1(-2)}{2 - 1}, \frac{2(-3) - 1(1)}{2 - 1} \right) = (2 + 2, -6 - 1) = (4, -7) \]
Step 3: Formulating the slope equation.
Any line through \( (4, -7) \) with slope \( m \) is \( y + 7 = m(x - 4) \implies mx - y - (4m + 7) = 0 \). Setting its distance from \( C_2(1, -3) \) equal to its radius \( r_2 = 2 \): \[ \frac{|m(1) - (-3) - 4m - 7|}{\sqrt{m^2 + 1}} = 2 \implies \frac{|-3m - 4|}{\sqrt{m^2 + 1}} = 2 \] Squaring both sides: \[ (3m + 4)^2 = 4(m^2 + 1) \implies 9m^2 + 24m + 16 = 4m^2 + 4 \implies 5m^2 + 24m + 12 = 0 \] The sum of the slopes (roots of this quadratic equation) is given by \( -\frac{\text{Coeff of } m}{\text{Coeff of } m^2} \): \[ m_1 + m_2 = -\frac{24}{5} \] This precisely matches Option (B).
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