Question

Let \( \theta \) be the angle between the circles \( x^{2}+y^{2}-4x+2fy-f=0 \) and \( x^{2}+y^{2}+2fx-4y-f=0 \). If \( \cos\theta=\frac{9}{16} \) and \( f\in\mathbb{Z} \), then the distance between the centres of these circles is

• A. \( \frac{10\sqrt{2}}{3} \)
• B. \( 5\sqrt{2} \)
• C. 5
• D. 13

Hint:

When both circles have equal radii, always reduce the angle formula to \( \cos\theta = 1 - \frac{d^2}{2r^2} \). It simplifies algebra drastically and avoids long expansions.

Answer 1

The Correct Option is B

Step 1: Find the centres and radii.
Circle 1 $x^2+y^2-4x+2fy-f=0$ has centre $C_1(2,-f)$ and $r_1^2=4+f^2+f$. Circle 2 $x^2+y^2+2fx-4y-f=0$ has centre $C_2(-f,2)$ and $r_2^2=f^2+4+f$. Notice $r_1=r_2=r$.
Step 2: Find the squared distance between centres.
\[ d^2=(2-(-f))^2+(-f-2)^2=(f+2)^2+(f+2)^2=2(f+2)^2 \]
Step 3: Write the angle formula.
For two circles meeting at angle $\theta$, \[ \cos\theta=\frac{d^2-r_1^2-r_2^2}{2r_1r_2}=\frac{d^2-2r^2}{2r^2}. \]
Step 4: Substitute and simplify.
With $r^2=f^2+f+4$ and $d^2=2(f+2)^2$: \[ \cos\theta=\frac{2(f+2)^2-2(f^2+f+4)}{2(f^2+f+4)}=\frac{3f}{f^2+f+4} \]
Step 5: Use $\cos\theta=\frac{9}{16}$ and pick the integer.
\[ \frac{3f}{f^2+f+4}=\frac{9}{16}\implies 48f=9(f^2+f+4)\implies 9f^2-39f+36=0 \] Dividing by 3 gives $3f^2-13f+12=0$, whose roots are $f=3$ and $f=\tfrac{4}{3}$. Since $f\in\mathbb{Z}$, take $f=3$.
Step 6: Find the distance.
\[ d^2=2(3+2)^2=2(25)=50\implies d=\sqrt{50}=5\sqrt{2} \] \[ \boxed{5\sqrt{2}} \]