If two circles \(x^2 + y^2 - 4x - 2y - 4 = 0\) \& \((x+1)^2 + (y+4)^2 = r^2\) intersect at two distinct points and range of r \(\in (\alpha, \beta)\), then the value of \(\alpha\beta\) is :
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Remember the geometric conditions for the intersection of two circles with radii \(R_1, R_2\) and distance \(d\) between centers:
Intersect at 2 points: \(|R_1 - R_2|<d<R_1 + R_2\)
Touch externally: \(d = R_1 + R_2\)
Touch internally: \(d = |R_1 - R_2|\)
Concept:
Gravitational force is a vector. We resolve forces from the four corners. Opposite masses exert forces in opposite directions, so we can subtract their magnitudes to find the net component along that diagonal.
Step 1: Initial Force Calculation (\(F_0\)).
Opposite pairs:
- \(4m\) and \(2m\) (Vertical/Diagonal 1): Net \(\propto 4-2 = 2m\).
- \(3m\) and \(m\) (Horizontal/Diagonal 2): Net \(\propto 3-1 = 2m\).
Resultant \(F_0 \propto \sqrt{(2)^2 + (2)^2} = 2\sqrt{2}\).
Step 2: Final Force Calculation (\(F'\)).
Swap \(4m\) and \(3m\). New arrangement:
- Vertical/Diagonal 1 now has \(3m\) and \(2m\): Net \(\propto 3-2 = 1m\).
- Horizontal/Diagonal 2 now has \(4m\) and \(m\): Net \(\propto 4-1 = 3m\).
Resultant \(F' \propto \sqrt{(1)^2 + (3)^2} = \sqrt{10}\).
Step 3: Ratio Calculation.
\[ \frac{F_0}{F'} = \frac{2\sqrt{2}}{\sqrt{10}} = \frac{2\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{2}{\sqrt{5}} \]
Step 4: Identify \(\alpha\).
Comparing with \(\frac{\alpha}{\sqrt{5}}\), we get \(\alpha = 2\).
\[
\boxed{\alpha=2}
\]
