Home Class 11 Applied Maths Chapter 16: Circle and Parabola Exercise 16.1: Circle
Chapter 16 • Unit VII

Exercise 16.1: Circle

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

1. Fundamental Definition and Standard Form of a Circle

In analytical geometry, a circle is defined as the locus of a point that moves in a two-dimensional plane such that its distance from a fixed point (called the center) remains constant (called the radius).

Standard (Center-Radius) Equation of a Circle

For a circle with center $C(h, k)$ and constant radius $r > 0$, any arbitrary point $P(x, y)$ on its circumference satisfies the Euclidean distance condition:

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

Squaring both sides yields the standard equation:

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

If the center is positioned at the origin $O(0, 0)$, the standard equation simplifies to:

$$x^2 + y^2 = r^2$$

2. The General Equation of a Circle

Expanding the standard equation gives:

$$(x^2 - 2hx + h^2) + (y^2 - 2ky + k^2) = r^2 \implies x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$$

Substituting $g = -h$, $f = -k$, and $c = h^2 + k^2 - r^2$, we obtain the general second-degree equation of a circle:

General Equation and Properties
$$x^2 + y^2 + 2gx + 2fy + c = 0$$
  • Center: $(-g, -f) = \left(-\frac{1}{2}\text{coeff of } x, -\frac{1}{2}\text{coeff of } y\right)$
  • Radius: $r = \sqrt{g^2 + f^2 - c}$

Geometric Nature of the Locus:

  1. If $g^2 + f^2 - c > 0$: The equation represents a real circle with positive radius.
  2. If $g^2 + f^2 - c = 0$: The equation represents a point circle (radius is zero, locus is strictly the single point $(-g, -f)$).
  3. If $g^2 + f^2 - c < 0$: The radius is imaginary (no real coordinates satisfy the equation).
Key Algebraic Rule: Before identifying $g, f, c$, the coefficients of $x^2$ and $y^2$ must be equal and normalized to $1$. There must be no cross product term ($xy$). If the equation is given as $Ax^2 + Ay^2 + Bx + Cy + D = 0$, divide every term by $A$ first.

3. Circle Described on a Given Diameter

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be the endpoints of a diameter of a circle. For any point $P(x, y)$ on the circumference (other than $A$ or $B$), the angle $\angle APB$ inscribed in a semicircle is a right angle ($90^\circ$).

Therefore, the product of the slopes of line segments $AP$ and $BP$ equals $-1$:

$$\left(\frac{y - y_1}{x - x_1}\right) \cdot \left(\frac{y - y_2}{x - x_2}\right) = -1 \implies (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$
Diameter Form
$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

4. Intercepts Made by a Circle on Coordinate Axes

To determine the length of the chords intercepted by the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ on the coordinate axes:

  • X-axis Intercept: Setting $y = 0$ gives the quadratic $x^2 + 2gx + c = 0$. The roots $x_1, x_2$ represent the x-coordinates of the intersection points. The intercept length is $|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2} = 2\sqrt{g^2 - c}$.
  • Y-axis Intercept: Setting $x = 0$ gives $y^2 + 2fy + c = 0$. The intercept length is $2\sqrt{f^2 - c}$.
Tangency Condition to Axes:
  • The circle touches the x-axis if $g^2 - c = 0 \implies c = g^2$.
  • The circle touches the y-axis if $f^2 - c = 0 \implies c = f^2$.
  • The circle touches both coordinate axes in the first quadrant if $g = -r, f = -r$ and $c = r^2$.

5. Step-by-Step Worked Examples

Example 1: Finding Center and Radius from General Form

Problem: Find the center and radius of the circle given by $2x^2 + 2y^2 - 12x + 16y - 6 = 0$.

Solution:

  1. Normalize the equation by dividing all terms by the leading coefficient $2$: $$x^2 + y^2 - 6x + 8y - 3 = 0$$
  2. Compare with standard general form $x^2 + y^2 + 2gx + 2fy + c = 0$: $$2g = -6 \implies g = -3$$ $$2f = 8 \implies f = 4$$ $$c = -3$$
  3. Compute the coordinates of the center: $$\text{Center} = (-g, -f) = (3, -4)$$
  4. Calculate the radius: $$r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 4^2 - (-3)} = \sqrt{9 + 16 + 3} = \sqrt{28} = 2\sqrt{7} \approx 5.29\text{ units}$$
Example 2: Determining Circle Equation from Three Non-Collinear Points

Problem: Find the equation of the circle passing through the origin $(0, 0)$ and having intercepts $4$ and $6$ on the positive x-axis and positive y-axis, respectively.

Solution:

  1. The circle passes through $O(0, 0)$, $A(4, 0)$, and $B(0, 6)$.
  2. Because the coordinate axes are perpendicular ($\angle AOB = 90^\circ$), the line segment $AB$ connecting $(4, 0)$ and $(0, 6)$ is a diameter of the circle (Thales's Theorem).
  3. Applying the diameter form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$: $$(x - 4)(x - 0) + (y - 0)(y - 6) = 0$$ $$x(x - 4) + y(y - 6) = 0 \implies x^2 - 4x + y^2 - 6y = 0$$
  4. Rearranging into standard general form: $$x^2 + y^2 - 4x - 6y = 0$$ Here, center is $(2, 3)$ and radius is $\sqrt{2^2 + 3^2 - 0} = \sqrt{13}$.
Example 3: Real-World Case Study — Telecom Signal Coverage Zone

Problem: A 5G cellular transceiver tower is installed at geographical coordinates $(2, -1)$ (units in kilometers). Field engineering tests indicate an omnidirectional coverage radius of $5\text{ km}$.

(a) Formulate the equation representing the outer boundary of this cellular coverage cell.
(b) An automated industrial IoT warehouse is situated at $(5, 3)$. Verify whether this facility lies inside, outside, or on the signal coverage boundary.

Solution:

  1. Coverage Boundary Equation:
    The tower is at center $(h, k) = (2, -1)$ with radius $r = 5$. $$(x - 2)^2 + (y - (-1))^2 = 5^2 \implies (x - 2)^2 + (y + 1)^2 = 25$$ Expanding: $$x^2 - 4x + 4 + y^2 + 2y + 1 = 25 \implies x^2 + y^2 - 4x + 2y - 20 = 0$$
  2. Facility Location Evaluation:
    Compute the Euclidean distance $d$ from the tower $(2, -1)$ to the warehouse $(5, 3)$: $$d = \sqrt{(5 - 2)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ km}$$ Alternatively, substitute $(5, 3)$ into $S = x^2 + y^2 - 4x + 2y - 20$: $$S(5, 3) = 5^2 + 3^2 - 4(5) + 2(3) - 20 = 25 + 9 - 20 + 6 - 20 = 0$$ Since $d = r = 5\text{ km}$ ($S = 0$), the warehouse is located precisely on the outer perimeter of the cell coverage zone. It maintains optimal line-of-sight signal access.
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