The problem requires finding the product $\alpha\beta$, where $(\alpha, \beta)$ represents the interval of possible values for the radius $r$ of the first circle, such that it intersects the second circle at two distinct points.
Circle Intersection Condition
Two circles intersect at two distinct points if the distance ($d$) between their centers is strictly greater than the absolute difference of their radii ($|R_1 - R_2|$) and strictly less than the sum of their radii ($R_1 + R_2$).
Condition: $|R_1 - R_2| < d < R_1 + R_2$
Circle Properties
Circle 1: $(x + 1)^2 + (y + 4)^2 = r^2$
- Center $C_1 = (-1, -4)$
- Radius $R_1 = r$
Circle 2: $x^2 + y^2 - 4x - 2y - 4 = 0$
Complete the square to find the center and radius:
$(x^2 - 4x) + (y^2 - 2y) = 4$
$(x^2 - 4x + 4) + (y^2 - 2y + 1) = 4 + 4 + 1$
$(x - 2)^2 + (y - 1)^2 = 9$
- Center $C_2 = (2, 1)$
- Radius $R_2 = \sqrt{9} = 3$
Distance Between Centers
Calculate the distance $d$ between $C_1(-1, -4)$ and $C_2(2, 1)$:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$d = \sqrt{(2 - (-1))^2 + (1 - (-4))^2}$
$d = \sqrt{(3)^2 + (5)^2}$
$d = \sqrt{9 + 25} = \sqrt{34}$
Finding the Interval for r
Apply the intersection condition using $R_1=r$, $R_2=3$, and $d=\sqrt{34}$:
$|r - 3| < \sqrt{34} < r + 3$
This yields two inequalities:
- $\sqrt{34} < r + 3 \implies r > \sqrt{34} - 3$
- $|r - 3| < \sqrt{34}$
Solving the second inequality:
$-\sqrt{34} < r - 3 < \sqrt{34}$
$3 - \sqrt{34} < r < 3 + \sqrt{34}$
Combine the results. Since $r$ must be positive (as it's a radius), and $\sqrt{34} \approx 5.83$, we have $\sqrt{34} - 3 \approx 2.83 > 0$ and $3 - \sqrt{34} \approx -2.83 < 0$. Therefore, the condition $r > \sqrt{34} - 3$ is the tighter lower bound.
The combined interval for $r$ is:
$r \in (\sqrt{34} - 3, \sqrt{34} + 3)$
This interval is given as $(\alpha, \beta)$. So, $\alpha = \sqrt{34} - 3$ and $\beta = \sqrt{34} + 3$.
Calculating $\alpha\beta$
Calculate the product $\alpha\beta$:
$\alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3)$
Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:
$\alpha\beta = (\sqrt{34})^2 - (3)^2$
$\alpha\beta = 34 - 9$
$\alpha\beta = 25$
Similar Questions
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Important Questions from Coordinate Geometry
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