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

• A. \( \frac{24}{5} \)
• B. \( -\frac{24}{5} \)
• C. \( \frac{8}{3} \)
• D. \( -\frac{8}{3} \)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Find both centres and radii.
Circle 1 $x^2+y^2+4x-2y-11=0$ has centre $C_1(-2,1)$ and radius $r_1=\sqrt{4+1+11}=4$. Circle 2 $x^2+y^2-2x+6y+6=0$ has centre $C_2(1,-3)$ and radius $r_2=\sqrt{1+9-6}=2$.
Step 2: Locate the external centre of similitude.
All common tangents pass through this point. It divides $C_1C_2$ externally in the ratio $r_1:r_2=4:2=2:1$. \[ T=\left(\frac{2(1)-1(-2)}{2-1},\ \frac{2(-3)-1(1)}{2-1}\right)=(4,-7) \]
Step 3: Write a general tangent through $T$.
A line through $(4,-7)$ with slope $m$ is $y+7=m(x-4)$, i.e. $mx-y-(4m+7)=0$.
Step 4: Make its distance from $C_2$ equal $r_2$.
A tangent touches the circle, so its distance from $C_2(1,-3)$ equals $2$: \[ \frac{|m(1)-(-3)-4m-7|}{\sqrt{m^2+1}}=2\implies\frac{|-3m-4|}{\sqrt{m^2+1}}=2 \]
Step 5: Square and tidy up.
\[ (3m+4)^2=4(m^2+1)\implies 9m^2+24m+16=4m^2+4 \] \[ 5m^2+24m+12=0 \]
Step 6: Use the sum of roots.
The two tangent slopes are the roots, so their sum is $-\dfrac{24}{5}$. \[ \boxed{-\tfrac{24}{5}} \]