🎯 Key Points
- General circle equation x²+y²+2gx+2fy+c=0 has centre (-g,-f) and radius √(g²+f²-c) — sign flip on g,f is a common slip
- Tangent to x²+y²=r² at point (x₁,y₁): replace x²→xx₁ and y²→yy₁, giving xx₁+yy₁=r² — a fast substitution trick, not a separate derivation each time
- Cone: V=⅓πr²h (one-third of the corresponding cylinder); Sphere: V=4/3πr³, SA=4πr² — sphere's volume formula is literally 4× the cone formula with h=r
- Frustum = cone with the top sliced off; its volume (1/3)πh(R²+r²+Rr) reduces to the full cone formula when r=0
Three standard 3D solids and their key formulas: a cylinder's volume scales with the full height, a cone's volume is exactly one-third of the cylinder with the same base and height, and a sphere's volume and surface area both follow distinct r-based formulas worth memorising separately.
Circles & Mensuration
Mensuration deals with the measurement of geometric figures: their perimeters, areas, and volumes.
2D Shapes
- Rectangle: Area = l×b, Perimeter = 2(l+b)
- Square: Area = a², Perimeter = 4a, Diagonal = a√2
- Triangle: Area = ½ bh; Heron's: √[s(s-a)(s-b)(s-c)], s=(a+b+c)/2
- Equilateral triangle: Area = (√3/4)a²
- Parallelogram: Area = base × height
- Trapezium: Area = ½(a+b)×h (a,b are parallel sides)
- Circle: Area = πr², Circumference = 2πr
- Sector: Arc = (θ/360)×2πr; Area = (θ/360)×πr²
3D Shapes
- Cube: Volume = a³, Surface area = 6a², Diagonal = a√3
- Cuboid: Volume = l×b×h, SA = 2(lb+bh+hl), Diagonal = √(l²+b²+h²)
- Cylinder: Volume = πr²h, CSA = 2πrh, TSA = 2πr(r+h)
- Cone: Volume = ⅓πr²h, CSA = πrl (l=slant height), TSA = πr(r+l)
- Sphere: Volume = (4/3)πr³, Surface area = 4πr²
- Hemisphere: Volume = (2/3)πr³, CSA = 2πr², TSA = 3πr²
Circle Theorems
- Angle in semicircle = 90°
- Angles in same segment are equal
- Angle at centre = 2 × angle at circumference (same arc)
- Tangent is perpendicular to radius at point of contact
- Tangent lengths from external point are equal
- Alternate segment theorem
Worked Example: Centre and Radius from the General Equation
Find the centre and radius of the circle x^2 + y^2 - 4x + 6y - 12 = 0.
Comparing with x^2 + y^2 + 2gx + 2fy + c = 0: 2g = -4 so g = -2; 2f = 6 so f = 3; c = -12.
Centre = (-g, -f) = (2, -3). Radius = sqrt(g^2 + f^2 - c) = sqrt(4 + 9 + 12) = sqrt(25) = 5.
Circle Through Diameter Endpoints
If A(x1, y1) and B(x2, y2) are the two endpoints of a diameter, the equation of the circle is (x - x1)(x - x2) + (y - y1)(y - y2) = 0.
Worked example: For A(2, 3) and B(8, 11): (x-2)(x-8) + (y-3)(y-11) = 0, which expands to x^2 + y^2 - 10x - 14y + 49 = 0. Its centre (5, 7) is the midpoint of AB, and its radius works out to 5, exactly half the diameter length AB = sqrt(6^2 + 8^2) = 10.
Tangent and Normal to a Circle
- The tangent to a circle x^2 + y^2 = r^2 at the point (x1, y1) on the circle is xx1 + yy1 = r^2 (replace x^2 by xx1 and y^2 by yy1).
- The normal at any point on a circle always passes through the centre, since the radius is perpendicular to the tangent at the point of contact.
- Worked example: For the circle x^2 + y^2 = 25 and the point P(3, 4) (which lies on the circle since 9 + 16 = 25), the tangent is 3x + 4y = 25. The normal passes through the centre (0,0) and P, giving the line 4x - 3y = 0.
Mensuration: Frustum of a Cone
A frustum is the solid left after cutting off the top of a cone with a plane parallel to its base. For a frustum of height h with radii R (bottom) and r (top):
- Volume = (1/3)πh(R^2 + r^2 + Rr)
- Curved surface area = πl(R + r), where slant height l = sqrt(h^2 + (R-r)^2)
- Total surface area = πl(R + r) + πR^2 + πr^2
Position of a Point Relative to a Circle
For the circle S: x² + y² + 2gx + 2fy + c = 0 and a point P(x₁,y₁), evaluate S₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c (substitute the point into the left side). Then:
- S₁ > 0 → P lies outside the circle
- S₁ = 0 → P lies on the circle
- S₁ < 0 → P lies inside the circle
Equivalently, compare the distance of P from the centre with the radius: farther than r means outside, equal means on, nearer means inside.
Circle Through Three Points
Exactly one circle passes through three non-collinear points. To find it, substitute each point into the general form x² + y² + 2gx + 2fy + c = 0. This gives three linear equations in the unknowns g, f, c; solving them determines the circle.
Worked example: the circle through (0,0), (6,0), (0,8). Substituting (0,0) gives c = 0. Substituting (6,0): 36 + 12g = 0 → g = -3. Substituting (0,8): 64 + 16f = 0 → f = -4. So the circle is x² + y² - 6x - 8y = 0, with centre (3,4) and radius 5.
Intersection of a Line and a Circle
Let d be the perpendicular distance from the centre of a circle of radius r to a given line. Comparing d with r decides how the line meets the circle:
- d < r → the line is a secant and cuts the circle at two distinct points
- d = r → the line is a tangent and touches the circle at exactly one point
- d > r → the line misses the circle (no common point)
Length of the chord cut off by a secant = 2√(r² - d²). The condition for the line y = mx + c to be tangent to x² + y² = r² is c² = r²(1 + m²).
Family of Circles and the Radical Axis
- Through a circle and a line: S + λL = 0 represents the family of circles passing through the intersection points of the circle S = 0 and the line L = 0, for a parameter λ.
- Through two circles: S₁ + λS₂ = 0 (λ ≠ -1) represents circles through the common points of S₁ = 0 and S₂ = 0.
- Radical axis: setting λ = -1 above eliminates the x² and y² terms and gives S₁ - S₂ = 0, a straight line called the radical axis. For two intersecting circles it is the common chord; the radical axis is always perpendicular to the line joining the two centres.
🚀 JEE Advanced Edge
Condition for a line to be tangent to a circle: A line touches the circle x²+y²=r² exactly when the perpendicular distance from the centre to the line equals the radius — this single distance-equals-radius condition replaces solving the line-circle system and checking for a repeated root (equal discriminant=0), and is the faster route to tangency problems where the line's equation has an unknown parameter to solve for.
Why the frustum formula generalizes both the cone and cylinder: Setting r=0 in V=(1/3)πh(R²+r²+Rr) gives (1/3)πR²h, the cone formula; setting r=R gives πR²h (since R²+R²+R²=3R², times 1/3 gives R²), the cylinder formula — recognizing the frustum as the general case helps verify a derived formula is dimensionally and structurally sound by checking it collapses correctly at the boundary cases.
Worked problem: Find the length of the tangent from the external point (5,0) to the circle x²+y²=9. Approach: Length of tangent from external point (x₁,y₁) to circle x²+y²=r² is √(x₁²+y₁²-r²). Here: √(25+0-9) = √16 = 4.