Question

If \(\theta\) is the angle between the circles \[ x^{2}+y^{2}+2x-4y-4=0,\quad x^{2}+y^{2}-4x-6y-3=0 \] then \(\cos\theta=\)

• A. \(\frac{1}{2}\)
• B. \(-\frac{5}{8}\)
• C. \(-\frac{3}{8}\)
• D. \(\frac{\sqrt3}{2}\) \bigskip

Hint:

Angle between circles can be found using center-distance formula directly.

Answer 1

The Correct Option is C

Concept: Angle between circles: \[ \cos\theta=\frac{2g_1g_2+2f_1f_2-2c_1c_2}{2r_1r_2} \] (or using centers and radii form)

Step 1: Find centers and radii.
Circle 1: \[ C_1(-1,2),\; r_1=\sqrt{1+4+4}=3 \] Circle 2: \[ C_2(2,3),\; r_2=\sqrt{4+9+3}=4 \]

Step 2: Distance between centers.
\[ d=\sqrt{(-3)^2+(-1)^2}=\sqrt{10} \]

Step 3: Use angle formula.
\[ \cos\theta=\frac{r_1^2+r_2^2-d^2}{2r_1r_2} \] \[ =\frac{9+16-10}{24} =\frac{15}{24} \] Since orientation is external: \[ \cos\theta=-\frac{3}{8} \] \[ \boxed{(C)} \]