Concept of Conic Section and class 10 Circle full CDC chapter

Add. Mathematics

Grade 10

Fully Unlocked

Lesson Conic section & Circle By: Ghanashyam Adhikari

Chapter Navigation

Interactive Simulation & Animation

Conic Sections (Circle, Parabola, Ellipse, Hyperbola)

Cutting Angle (\(\theta\)): 90.0°

Select Conic Section Type:

Observe how the cutting plane angle forms circle, ellipse, parabola, or hyperbola.

Coordinate Plane: Circle Graph

Standard Equation: \(x^2 + y^2 = r^2\)

Activity 1 • Real-Life Applications

Conics Around Us

Study the figures below and identify their geometric properties:

Figure (a): Clock Face

Shape: Circle

The visible outer part of the clock is a circle, and the distance from its centre to the circumference is equal in all directions (\(r\)).

Figure (b): Planetary Orbits

Shape: Ellipse

Planets orbiting the Sun follow elliptical paths rather than circular ones, having two foci instead of a single equidistant centre.

Figure (c): Nepal Telecom Dish

Shape: Parabola

Satellite dishes reflect incoming parallel signals to a single focal point, utilizing parabolic reflective properties.

Figure (d): Sand Hourglass

Shape: Hyperbola (Double Cone)

The sand hourglass formed by joining two cones vertex-to-vertex illustrates double-napped conical intersection.

Section 8.4.1 • Equation of Circle

Derivation of Standard Equations & Required Facts

Strictly adhering to CDC Grade 10 Additional Mathematics curriculum

a. Equation of a circle with centre at origin \((0, 0)\) and radius \(r\)

In the given figure, the centre of the circle is at the origin \(O(0, 0)\). Let \(P(x, y)\) be a point on the circumference. The distance \(OP\) from the centre \(O\) to the point \(P(x, y)\) on the circumference is the radius of the circle, where \(OP = r\).

Assume \(O(0, 0) = (x_1, y_1)\) and \(P(x, y) = (x_2, y_2)\).

Now, by the distance formula:

\[OP = d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] \[\text{or, } r = \sqrt{(x - 0)^2 + (y - 0)^2}\] \[\text{or, } r = \sqrt{(x)^2 + (y)^2}\] \[\text{or, } r = \sqrt{x^2 + y^2}\]

Squaring on both sides: \(x^2 + y^2 = r^2\)

b. Equation of a circle with centre \((h, k)\) and radius \(r\)

In the given figure, point \(C(h, k)\) is the centre of the circle. Let \(P(x, y)\) be a point on the circumference. The distance \(CP\) from the centre \(C\) to the point \(P(x, y)\) on the circumference is the radius of the circle, where \(CP = r\).

Now, assume \(C(h, k) = (x_1, y_1)\) and \(P(x, y) = (x_2, y_2)\).

Now, by the distance formula: \(CP^2 = r^2\)

\[\text{or, } (x_2 - x_1)^2 + (y_2 - y_1)^2 = r^2\] \[\text{or, } (x - h)^2 + (y - k)^2 = r^2\]

Therefore, the equation of a circle with centre \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\).

c. Equation of a circle having the endpoints of a diameter as \((x_1, y_1)\) and \((x_2, y_2)\)

In the adjacent figure, point \(C\) is the centre of the circle. The endpoints of the diameter \(AB\) are \(A(x_1, y_1)\) and \(B(x_2, y_2)\). Let \(P(x, y)\) be a point on the circumference. Join \(PA\) and \(PB\).

\(\angle APB\) is an angle inscribed in a semicircle, therefore \(\angle APB = 90^\circ\).

Now, for the slope of line \(AP\), assume \(A(x_1, y_1) = (x_1, y_1)\) and \(P(x, y) = (x_2, y_2)\).

\[(m_1) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}\] \[\text{Similarly, the slope of } BP (m_2) = \frac{y - y_2}{x - x_2}\]

We know that \(AP\) and \(BP\) are perpendicular to each other, so: \(m_1 \times m_2 = -1\)

\[\text{or, } \frac{y - y_1}{x - x_1} \times \frac{y - y_2}{x - x_2} = -1\] \[\text{or, } \frac{(y - y_1)(y - y_2)}{(x - x_1)(x - x_2)} = -1\] \[\text{or, } (y - y_1)(y - y_2) = -(x - x_1)(x - x_2)\] \[\text{or, } (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\]

Therefore, equation of circle having diameter with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is \((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\).

d. General form of equation of circle

The equation of circle with centre \((h, k)\) and radius \((r)\) is \((x - h)^2 + (y - k)^2 = r^2\)

\[\text{or, } x^2 - 2xh + h^2 + y^2 - 2yk + k^2 = r^2\] \[\text{or, } x^2 + y^2 + 2(-h)x + 2(-k)y + h^2 + k^2 - r^2 = 0\] \[\text{or, } x^2 + y^2 + 2gx + 2fy + c = 0 \quad \text{(called general form of equation of circle)}\]

Where, \(g = (-h), f = (-k) \text{ and } c = h^2 + k^2 - r^2\)

Where, \(h = -g, k = -f \text{ and } r^2 = h^2 + k^2 - c, r = \sqrt{h^2 + k^2 - c}\)

Therefore, centre of circle \((h, k) = (-g, -f)\) and radius \((r) = \sqrt{g^2 + f^2 - c}\) (or \(\sqrt{h^2 + k^2 - c}\)).

The general equation \(x^2 + y^2 + 2gx + 2fy + c = 0\) has following characteristics:

• The degree of \(x\) and \(y\) is 2.
• Coefficient of \(x^2 =\) Coefficient of \(y^2\).

Required Facts (Summary Table)

S.N.	Condition of Circle	Equation of Circles
1	Centre at origin \((0, 0)\)	\(x^2 + y^2 = r^2\)
2	Centre at \((h, k)\)	\((x - h)^2 + (y - k)^2 = r^2\)
3	The endpoints of diameter are given \((x_1, y_1), (x_2, y_2)\)	\((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\)
4	When circle touches X-axis (\(r = k\))	\((x - h)^2 + (y - k)^2 = k^2 \quad \text{or} \quad (x - h)^2 + (y - r)^2 = r^2\)
5	When circle touches Y-axis (\(r = h\))	\((x - h)^2 + (y - k)^2 = h^2 \quad \text{or} \quad (x - r)^2 + (y - k)^2 = r^2\)
6	When circle touches both the axes (\(r = h = k\))	\((x - h)^2 + (y - h)^2 = h^2 \text{ or } (x - k)^2 + (y - k)^2 = k^2 \text{ or } (x - r)^2 + (y - r)^2 = r^2\)

Thought Provoking Question: What happens when a circle touches both the axes and lies on second or third or fourth quadrant? Discuss.

Worked Examples 1 to 7

Step-by-Step Textbook Solutions

Solution by Ghanashyam Adhikari

Example 1

Find the equation of the circle with centre \((1, -5)\) and radius \(4\) units.

Solution: Here,

Centre of the circle \((h, k) = (1, -5)\) and radius \((r) = 4\) units.

According to the formula, the equation of the circle is:

\((x - h)^2 + (y - k)^2 = r^2\)

\(\implies (x - 1)^2 + (y + 5)^2 = 4^2\)

\(\implies x^2 - 2 \cdot x \cdot 1 + (1)^2 + y^2 - 2 \cdot y \cdot (-5) + (-5)^2 = 16\)

\(\implies x^2 - 2x + 1 + y^2 + 10y + 25 = 16\)

\(\implies x^2 + y^2 - 2x + 10y + 1 + 25 - 16 = 0\)

\(\implies x^2 + y^2 - 2x + 10y + 10 = 0\) is the required equation of the circle.

Example 2

If the endpoints of a diameter of a circle are \((2, -4)\) and \((-3, 7)\), find the equation of the circle.

Solution: Here,

Let the endpoints of the diameter be \((2, -4) = (x_1, y_1)\) and \((-3, 7) = (x_2, y_2)\).

The equation of the circle having endpoints \((x_1, y_1)\) and \((x_2, y_2)\) of its diameter is:

\((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\)

\(\implies (x - 2)(x - (-3)) + (y - (-4))(y - 7) = 0\)

\(\implies (x - 2)(x + 3) + (y + 4)(y - 7) = 0\)

\(\implies x(x + 3) - 2(x + 3) + y(y - 7) + 4(y - 7) = 0\)

\(\implies x^2 + 3x - 2x - 6 + y^2 - 7y + 4y - 28 = 0\)

\(\implies x^2 + y^2 + x - 3y - 34 = 0\) is required equation of the circle.

Example 3

If the equation of a circle is \(9x^2 + 9y^2 - 36x + 6y - 107 = 0\), find the coordinates of its centre and its diameter.

Solution: Here,

Given equation of the circle: \(9x^2 + 9y^2 - 36x + 6y - 107 = 0\)

Divide both sides of the equation by \(9\):

\(x^2 + y^2 - 4x + \frac{2}{3}y - \frac{107}{9} = 0 \quad \text{....(i)}\)

Comparing equation (i) with the general form \(x^2 + y^2 + 2gx + 2fy + c = 0\), we get:

\(2g = -4 \implies g = -2\)

\(2f = \frac{2}{3} \implies f = \frac{1}{3}\) and \(c = -\frac{107}{9}\)

According to the formula, the centre \((h, k) = (-g, -f) = (2, -\frac{1}{3})\)

The radius \((r) = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (\frac{1}{3})^2 - (-\frac{107}{9})} = \sqrt{4 + \frac{1}{9} + \frac{107}{9}} = \sqrt{\frac{36 + 1 + 107}{9}} = \sqrt{\frac{144}{9}} = 4\)

Diameter \((d) = 2r = 2 \times 4 = 8\) units.

Therefore, the coordinates of the centre are \((2, -\frac{1}{3})\) and the diameter is \(8\) units.

Example 4

Find the equation of the circle that lies entirely in the third quadrant, touches both axes, and has a radius of \(5\) units.

Solution: Here, Radius \((r) = 5\) units.

The circle touches both the axes and lies completely in the third quadrant.

So, \((h, k) = (-r, -r) = (-5, -5)\)

By the formula, \((x - h)^2 + (y - k)^2 = r^2 \implies (x - (-5))^2 + (y - (-5))^2 = 5^2\)

\(\implies (x + 5)^2 + (y + 5)^2 = 25\)

\(\implies x^2 + 10x + 25 + y^2 + 10y + 25 = 25\)

\(\implies x^2 + y^2 + 10x + 10y + 25 = 0\) is the required equation.

Example 5

Find the equation of the circle having the diameter with \(x\)-intercept \(12\) units and \(y\)-intercept \(8\) units.

Solution: Here, \(x\)-intercept \(= 12\), \(y\)-intercept \(= 8\).

The circle passes through the origin. Therefore, \(AB\) is the diameter where \(A(12, 0)\) and \(B(0, 8)\).

Using diameter endpoints formula:

\((x - 12)(x - 0) + (y - 0)(y - 8) = 0\)

\(\implies (x - 12)x + y(y - 8) = 0\)

\(\implies x^2 + y^2 - 12x - 8y = 0\) is the required equation.

Example 6

Find the equation of the circle that passes through points \(A(2, 3)\) and \(B(5, 4)\) and whose centre lies on the straight line \(2x + 3y = 7\).

Solution: Here, let centre be \((h, k)\).

Since \((h, k)\) lies on \(2x + 3y = 7\), we have \(2h + 3k = 7 \implies h = \frac{7 - 3k}{2} \quad \text{....(i)}\)

Again, \(PA = PB\) (radii of same circle):

\((h - 2)^2 + (k - 3)^2 = (h - 5)^2 + (k - 4)^2\)

Solving gives \(k = -1\). Substituting in (i) gives \(h = 5\).

Centre \((5, -1)\), Radius \(r = 5\). Equation: \(x^2 + y^2 - 10x + 2y + 1 = 0\).

Example 7

Find the equation of the circle concentric with the circle \(x^2 + y^2 - 8x + 6y - 5 = 0\) and passing through the point \((-2, -7)\).

Solution: Here, given circle equation: \(x^2 + y^2 - 8x + 6y - 5 = 0\)

Comparing with general form: \(2g = -8 \implies g = -4\), \(2f = 6 \implies f = 3\), \(c = -5\)

Centre of given circle \((h, k) = (-g, -f) = (4, -3)\). Concentric circle has the same centre \((4, -3)\).

Radius \(r =\) distance from \((4, -3)\) to \((-2, -7) = \sqrt{(-2 - 4)^2 + (-7 - (-3))^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\)

Equation: \((x - 4)^2 + (y + 3)^2 = (\sqrt{52})^2 \implies x^2 + y^2 - 8x + 6y - 27 = 0\).

Chapter Assessment • Gamified Practice

Interactive Quizzes

Score: 0 / 8

Textbook Exercise Solutions

Exercise Questions (Q3 to Q14)

Official CDC Nepal Model Grid

Examination Section (Full Model Paper)

Total Marks: 50 5 (2m) + 5 (3m) + 5 (5m)

Instructions for Students

Answer all 15 questions below. You can use the Mathematics Equation Editor or upload handwritten solution images (JPG/PNG/WEBP). Click Submit Exam for Auto-Marking when finished.

Math Equation Editor

Upload Handwritten Solution (JPG/PNG)

CDC Model Question Paper (Full Conic Sections & Circles)

Group A: Short Answer Questions (5 Questions × 2 Marks = 10 Marks)

1. Find the centre and radius of the circle given by equation \(x^2 + y^2 - 6x + 8y - 11 = 0\).
2. Find the equation of the circle with centre \((2, -3)\) and radius \(5\) units.
3. Find the equation of the circle if endpoints of its diameter are \((2, -4)\) and \((-3, 7)\).
4. Under what condition is a parabola formed when a plane cuts a right circular cone?
5. Find the equation of the circle touching both axes, lying in the third quadrant with radius \(4\) units.

Group B: Medium Answer Questions (5 Questions × 3 Marks = 15 Marks)

6. Find the equation of the circle passing through points \(M(-2, 0)\) and \(N(0, -2)\) whose centre lies on the line \(2x - 3y + 1 = 0\).
7. Find the equation of the circle concentric with \(x^2 + y^2 - 8x + 6y - 5 = 0\) and passing through \((-2, -7)\).
8. Find the equation of the circle having \(x\)-intercept \(12\) units and \(y\)-intercept \(8\) units, passing through the origin.
9. Explain with a neat diagram the semi-vertical angle \(\alpha\) and cutting angle \(\theta\) for forming an ellipse.
10. Find the centre and radius of the circle represented by \(9x^2 + 9y^2 - 36x + 6y - 107 = 0\).

Group C: Long Answer / Higher-Order Questions (5 Questions × 5 Marks = 25 Marks)

11. Find the equation of the circle passing through three points \((0, 0)\), \((4, 0)\), and \((0, 2)\). Show step-by-step general form substitution.
12. A circular flower garden has centre \((2, -1)\) and radius \(7\) metres. Write its equation and determine whether a student at \((8, -1)\) is inside, outside, or on the boundary.
13. Find the equation of the circle that passes through points \(A(2, 3)\) and \(B(5, 4)\) and whose centre lies on the straight line \(2x + 3y = 7\).
14. Derive the standard equation of a circle with centre at the origin \((0, 0)\) and radius \(r\) using coordinate geometry and the distance formula.
15. Explain double-napped right circular cones and describe how circles, ellipses, parabolas, and hyperbolas are generated by varying the cutting plane angle \(\theta\) relative to the semi-vertical angle \(\alpha\).

Pen:

Size: 1px 2px 3px 4px