Let the set of all values of r for which circles (x+1)² + (y+4)² = r² and x² + y² – 4x – 2y – 4 = 0 intersect at two distinct points be interval (α,β). Then αβ is equal to

When two circles with centers C₁ and C₂ and radii r₁ and r₂ are given, nature of intersection depends on distance d between centers. If d > r₁ + r₂, circles are completely separate with no intersection points. If d = r₁ + r₂, they touch externally at exactly one point (external tangency). If |r₁ – r₂| < d < r₁ + r₂, they intersect at exactly two distinct points - this is most common case tested in JEE. If d = |r₁ - r₂|, they touch internally at one point (internal tangency). If d < |r₁ - r₂|, one circle lies completely inside other without touching. These geometric conditions translate to algebraic inequalities. Understanding this helps visualize problem before computation. Two-point intersection requires distance be "just right" - not too small (nested circles) and not too large (separated circles). In JEE problems, you're often given interval for one radius and asked to find properties of endpoints. Remember: |r₁ - r₂| ensures we take positive difference regardless of which radius is larger. This is crucial when r varies as in our problem.

A circle has two main equation forms. Standard form: $(x-h)^2 + (y-k)^2 = r^2$ where (h,k) is center and r is radius. Immediately shows geometric properties. Expanding gives general form: $x^2 + y^2 + 2gx + 2fy + c = 0$. To convert general to standard, complete the square. From general form, center is (-g, -f) and radius is $\sqrt{g^2 + f^2 – c}$. For real circle to exist, we need $g^2 + f^2 – c > 0$. If this equals zero, equation represents a point circle (radius = 0). If negative, no real circle exists. Important: in general form, coefficients of x² and y² must be equal (both 1 after dividing) and there should be no xy term, otherwise it’s not a circle but some other conic. When working with two circles, converting both to same form simplifies distance calculations. Standard form is better for finding centers quickly. General form is better for algebraic manipulations like finding common chord equation or radical axis. Practice converting between forms quickly for JEE speed.

Distance between two points (x₁,y₁) and (x₂,y₂) in plane is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Derived from Pythagorean theorem. In 3D, extends to $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. Applications include: finding length of line segments, verifying if three points are collinear (sum of two distances equals third), checking if point lies on circle (distance from center equals radius), finding locus of points equidistant from two given points (perpendicular bisector). In circle problems, distance between centers determines intersection nature. Common optimization: use squared distance $d^2$ to avoid square roots in intermediate steps, taking square root only at end. For JEE, memorize common Pythagorean triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25) for quick mental calculation. Section formula for point dividing line segment in ratio m:n is directly related to weighted distance. Distance formula is fundamental in coordinate geometry – master it thoroughly.

Difference of squares: $(a-b)(a+b) = a^2 – b^2$ is most useful identity for JEE. Allows instant multiplication without expanding all terms. In our problem, $(\sqrt{34}-3)(\sqrt{34}+3) = 34 – 9 = 25$ saves significant time. Other key identities: $(a+b)^2 = a^2 + 2ab + b^2$, $(a-b)^2 = a^2 – 2ab + b^2$, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$. For sums and differences of cubes: $a^3 + b^3 = (a+b)(a^2-ab+b^2)$, $a^3 – b^3 = (a-b)(a^2+ab+b^2)$. Component identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$. Also useful: $a^2 + b^2 = (a+b)^2 – 2ab$. Recognize patterns like $x^2 – 7 = (x-\sqrt{7})(x+\sqrt{7})$ for any number. These speed up factorization, simplification, and solving equations. Practice identifying which identity applies quickly – this comes with extensive problem practice. In competitive exams, recognizing the right identity can save 30-60 seconds per problem, which accumulates significantly over full paper.