Mathematics Class 11
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Sections of a Cone and the Circle

🎓 Class 11 Mathematics CBSE Theory Ch 10 — Conic Sections ⏱ ~30 min
🌐 ભાષા:

આ MCQ મોડ્યુલ આના પર આધારિત છે: Sections of a Cone and the Circle

આ મૂલ્યાંકન આના પર આધારિત હશે: Sections of a Cone and the Circle
Class 11 સ્તર, Coordinate Geometry માં, Advanced કઠિનતા સાથે લક્ષ્ય.

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

10.1 Introduction

In the previous chapter we studied straight lines. In this chapter we study curves obtained as the plane sections of a double right-circular conei. Depending on the angle between the cutting plane and the axis of the cone, we obtain four non-degenerate curves — the circle, ellipse, parabola and hyperbola — collectively called conic sections or simply conics. These curves are central to astronomy (planetary orbits), engineering (headlight reflectors, arches), and mechanics (projectile paths).

Apollonius of Perga
(c. 262 – 190 BCE)
Greek geometer. His 8-book treatise Conics named the parabola, ellipse and hyperbola and developed almost all their classical properties.

10.2 Sections of a Cone

Take a fixed vertical line \(l\) and a line \(m\) intersecting it at a point \(V\) at a constant angle \(\alpha\) (\(0<\alpha<90°\)). Rotate \(m\) around \(l\) keeping the angle fixed. The surface swept out is a double right-circular cone with vertex \(V\), axis \(l\) and semi-vertical angle \(\alpha\). The two half-cones above and below \(V\) are called nappes. Any line on the surface through \(V\) is a generatori.

l (axis) V α upper nappe lower nappe
Fig 10.1 — A double right-circular cone with vertex \(V\), axis \(l\), semi-vertical angle \(\alpha\).

10.2.1 The four conic sections

Let \(\beta\) be the acute angle the intersecting plane makes with the axis \(l\) (and the plane does not pass through \(V\)). Then the intersection with the cone is:

  • A circle when the plane is perpendicular to the axis (\(\beta=90°\)).
  • An ellipse when \(\alpha<\beta<90°\) (oblique plane cutting one nappe only).
  • A parabola when \(\beta=\alpha\) (plane parallel to a generator).
  • A hyperbola when \(0\le\beta<\alpha\) (plane cuts both nappes).
Circle Ellipse Parabola Hyperbola
Fig 10.2 — The four non-degenerate conic sections obtained from a double cone.

10.2.2 Degenerate conics

When the cutting plane passes through the vertex \(V\), the section degenerates to: a point (\(\beta>\alpha\)); a single line (\(\beta=\alpha\), plane tangent to cone along a generator); or a pair of intersecting lines (\(\beta<\alpha\)).

10.3 Circle

Definition
A circlei is the set of all points in a plane that are equidistant from a fixed point. The fixed point is the centre and the common distance is the radius.

10.3.1 Standard equation

Let the centre be \(C(h,k)\) and radius \(r>0\). A point \(P(x,y)\) lies on the circle iff \(CP=r\):

\(\sqrt{(x-h)^2+(y-k)^2}=r\ \Longleftrightarrow\ (x-h)^2+(y-k)^2=r^2.\)

Expanding, we get the general second-degree form \(x^2+y^2+2gx+2fy+c=0\) with centre \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\) (provided \(g^2+f^2-c>0\)).

xy C(h,k) r P(x,y)
Fig 10.3 — Circle of centre \(C(h,k)\) and radius \(r\): \((x-h)^2+(y-k)^2=r^2\).
Example 1
Find the equation of the circle with centre \((-3,2)\) and radius 4.
\((x+3)^2+(y-2)^2=16\Rightarrow x^2+y^2+6x-4y-3=0.\)
Example 2
Find the centre and radius of \(x^2+y^2+8x+10y-8=0\).
Here \(2g=8,2f=10,c=-8\Rightarrow g=4,f=5\). Centre \((-4,-5)\). Radius \(=\sqrt{16+25+8}=\sqrt{49}=7.\)
Example 3 — Through three points
Find the equation of the circle passing through \((2,-2),(3,4),(-1,6)\).
Let circle be \(x^2+y^2+2gx+2fy+c=0\). Plug each point: \(4+4+4g-4f+c=0\), \(9+16+6g+8f+c=0\), \(1+36-2g+12f+c=0\). Simplify: (i) \(4g-4f+c=-8\); (ii) \(6g+8f+c=-25\); (iii) \(-2g+12f+c=-37\). (ii)-(i): \(2g+12f=-17\). (iii)-(i): \(-6g+16f=-29\). Solve: from first pair \(g=-\tfrac{17+12f}{2}\); substitute: \(-6(-\tfrac{17+12f}{2})+16f=-29\Rightarrow 3(17+12f)+16f=-29\Rightarrow 51+52f=-29\Rightarrow f=-\tfrac{80}{52}=-\tfrac{20}{13}\); then \(g=-\tfrac{17+12(-20/13)}{2}=-\tfrac{17-240/13}{2}=-\tfrac{(221-240)/13}{2}=\tfrac{19}{26}\). Using (i): \(c=-8-4g+4f=-8-\tfrac{38}{13}+(-\tfrac{80}{13})=-\tfrac{104+38+80}{13}=-\tfrac{222}{13}\). Hence circle: \(x^2+y^2+\tfrac{19}{13}x-\tfrac{40}{13}y-\tfrac{222}{13}=0\), i.e. \(13(x^2+y^2)+19x-40y-222=0\).
Example 4
Does \((2,1)\) lie inside, on, or outside the circle \(x^2+y^2-4x+6y-12=0\)?
Substitute: \(4+1-8+6-12=-9<0\). The LHS (after moving RHS) evaluated at the point is negative \(\Rightarrow\) inside the circle.
Activity 10.1 — Slicing a paper cone
Predict: Roll a paper into a cone. If you slice it perpendicular to its axis, exactly parallel to its slant, and tilted steeply, what three curves will you expose?
  1. Make the three slices on a paper cone and trace the boundaries.
  2. Classify each curve as circle / ellipse / parabola / hyperbola using the angle rule \(\alpha\) vs \(\beta\).
  3. With a stack of two paper cones tip-to-tip, try a slice parallel to the axis — confirm you get a hyperbola (two branches).
  4. Record which conic has the smallest eccentricity and which the largest (preview for later sections).
Insight: Perpendicular → circle (e=0). Parallel to generator → parabola (e=1). Parallel to axis → hyperbola (e>1). Oblique → ellipse (0

In-text Exercises on Circles

Q1. Find the equation of the circle with centre \((0,2)\) and radius 2.
\(x^2+(y-2)^2=4\Rightarrow x^2+y^2-4y=0.\)
Q2. Find the centre and radius of \(x^2+y^2-4x-8y-45=0\).
Centre \((2,4)\). \(r=\sqrt{4+16+45}=\sqrt{65}.\)
Q3. Find the equation of the circle passing through \((4,1)\) and \((6,5)\) whose centre is on the line \(4x+y=16\).
Let centre \((h,k)\) with \(4h+k=16\). Equidistance: \((h-4)^2+(k-1)^2=(h-6)^2+(k-5)^2\). Expand: \(-8h+16-2k+1=-12h+36-10k+25\Rightarrow 4h+8k=44\Rightarrow h+2k=11\). Solve with \(4h+k=16\): multiply first by 4: \(4h+8k=44\); subtract: \(7k=28\Rightarrow k=4\). Then \(h=3\). Radius\(^2=(3-4)^2+(4-1)^2=10\). Circle: \((x-3)^2+(y-4)^2=10.\)
Q4. Find the equation of the circle with centre \((1,1)\) which passes through \((-2,5)\).
\(r^2=(1+2)^2+(1-5)^2=9+16=25\). \((x-1)^2+(y-1)^2=25.\)
Competency-Based Questions — Circular Boundaries
A stadium's circular athletic track has centre at \(O(3,-2)\) (coordinates in metres) and the spectator stand is known to touch the track at the point \(A(7,1)\).
Q1. Determine the radius and equation of the track.
L3 Apply
\(r=\sqrt{(7-3)^2+(1+2)^2}=\sqrt{16+9}=5\) m. Equation: \((x-3)^2+(y+2)^2=25.\)
Q2. Check whether the point \(B(0,2)\) lies inside, on, or outside the track, showing work.
L4 Analyse
\(OB=\sqrt{(0-3)^2+(2+2)^2}=\sqrt{9+16}=5=r\). The point \(B\) lies exactly on the track.
Q3. Two lamp-posts stand at \(P(6,2)\) and \(Q(-1,3)\). Which of them can be used as a "corner post" sitting on the track's boundary?
L5 Evaluate
\(OP=\sqrt{9+16}=5\) — on the circle. \(OQ=\sqrt{16+25}=\sqrt{41}\ne 5\) — off the circle. Only \(P\) can serve as a corner post on the track.
Q4. Design a concentric "warm-up track" whose radius is 3 m greater than the main track and write its general equation \(x^2+y^2+2gx+2fy+c=0\).
L6 Create
New radius = 8 m, same centre \((3,-2)\): \((x-3)^2+(y+2)^2=64\Rightarrow x^2+y^2-6x+4y-51=0\). Here \(2g=-6,2f=4,c=-51\), giving \(g=-3,f=2,c=-51\).
Assertion–Reason Questions
Assertion (A): The equation \(x^2+y^2-4x+6y+25=0\) represents a real circle.
Reason (R): For \(x^2+y^2+2gx+2fy+c=0\) to be a real circle, \(g^2+f^2-c>0\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. Here \(g=-2,f=3,c=25\); \(g^2+f^2-c=4+9-25=-12<0\), so A is false. R is a correct general criterion.
Assertion (A): When a plane cuts a double cone parallel to a generator, the section is a parabola.
Reason (R): A plane parallel to a generator makes the same angle with the axis as the cone's semi-vertical angle \(\alpha\), i.e. \(\beta=\alpha\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. \(\beta=\alpha\) is precisely the parabola condition; R explains A.
Assertion (A): The general second-degree equation \(x^2+y^2+2gx+2fy+c=0\) always represents a circle.
Reason (R): In a circle, the coefficients of \(x^2\) and \(y^2\) are equal and there is no \(xy\)-term.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. A is false — it represents a circle only when \(g^2+f^2-c>0\); otherwise it's a point (=0) or imaginary (<0). R is correct structurally.

Frequently Asked Questions

How is a circle obtained from a cone?
When a plane cuts a double cone perpendicular to the axis, the intersection is a circle. A circle is a special case of an ellipse when the plane tilt is zero.
What is the equation of a circle with centre (h, k) and radius r?
The standard equation is (x - h) squared + (y - k) squared = r squared. If the centre is the origin, it becomes x squared + y squared = r squared.
What is the general equation of a circle?
The general form is x squared + y squared + 2 g x + 2 f y + c = 0, with centre (-g, -f) and radius root(g squared + f squared - c).
When does slicing a cone give a parabola?
When the slicing plane is parallel to a slant (generator) of the cone, the cross-section is a parabola.
When does slicing a cone give a hyperbola?
When the slicing plane cuts both parts of the double cone, the cross-section is a hyperbola, with two separate branches.
What are the real-life applications of conic sections?
Planetary orbits (ellipses), satellite dishes and car headlights (parabolas), cooling towers and sonic booms (hyperbolas), and wheels and clocks (circles).
AI ટ્યુટર
Mathematics Class 11
તૈયાર
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