Equation of a circle Knowledge-Based Study

Study notes

1. What is a Circle?

   In coordinate geometry, a circle is defined as the set of all points in a plane that are at a fixed distance, called the radius, from a fixed point called the center.

   Deriving the Equation

   Let the center of the circle be the point $C(h, k)$ and let the radius be $r$, where $r > 0$. If $P(x, y)$ is any arbitrary point lying on the circle, the distance between $C$ and $P$ must equal $r$.

   Using the Euclidean distance formula:

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

   Squaring both sides eliminates the radical and yields the fundamental relationship for all points on the circle:

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

   Standard (Center-Radius) Form

   The equation

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

   is known as the standard form of the equation of a circle.

   • Center: $(h, k)$
   • Radius: $r = \sqrt{r^2}$

   Special Case: Center at the Origin

   When the center is positioned at the origin $(0, 0)$, substituting $h = 0$ and $k = 0$ simplifies the equation to:

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

   Example

   Consider a circle with center $(3, -4)$ and radius $5$. Its standard equation is:

   $$(x - 3)^2 + (y - (-4))^2 = 5^2$$$$(x - 3)^2 + (y + 4)^2 = 25$$

   Note: Pay close attention to the signs. The expression $(y + 4)^2$ indicates that the $y$-coordinate of the center is $-4$, not $4$.

   Expanding to General Form

   Expanding the binomials in the standard equation $(x - h)^2 + (y - k)^2 = r^2$ gives:

   $$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2$$

   Rearranging the terms in descending degree order:

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

   By defining constants $D = -2h$, $E = -2k$, and $F = h^2 + k^2 - r^2$, we obtain the general form of a circle's equation:

   $$x^2 + y^2 + Dx + Ey + F = 0$$

   Key properties of the general form representing a circle in the Cartesian plane:

   • The coefficients of $x^2$ and $y^2$ are equal and non-zero (if multiplied by a constant $A$, it appears as $Ax^2 + Ay^2 + Dx + Ey + F = 0$).
   • There is no cross-product term involving $xy$.

   Completing the Square

   To identify the center and radius from an equation given in general form, we use the method of completing the square for both $x$ and $y$.

   Step-by-Step Procedure

   1. If the coefficients of $x^2$ and $y^2$ are greater than 1, divide the entire equation by that coefficient.
   2. Group the $x$-terms and $y$-terms together, and move the constant term to the right side.
   3. Add $\left(\frac{D}{2}\right)^2$ and $\left(\frac{E}{2}\right)^2$ to both sides of the equation.
   4. Factor each group into a perfect square: $(x - h)^2 + (y - k)^2 = M$.

   Analysis of the Constant $M$

   • If $M > 0$, the equation represents a real circle with radius $r = \sqrt{M}$.
   • If $M = 0$, the equation represents a single point $(h, k)$, often called a degenerate circle.
   • If $M < 0$, no real points satisfy the equation (an imaginary circle).

   Example 1: Finding Center and Radius

   Find the center and radius of the circle given by the equation:

   $$x^2 + y^2 - 6x + 8y - 11 = 0$$

   Solution:

   Group variables and move the constant:

   $$(x^2 - 6x) + (y^2 + 8y) = 11$$

   Complete the square for both groups:

   • For $x$: $\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$
   • For $y$: $\left(\frac{8}{2}\right)^2 = (4)^2 = 16$

   Add these values to both sides:

   $$(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16$$$$(x - 3)^2 + (y + 4)^2 = 36$$

   Therefore, the circle has:

   • Center: $(3, -4)$
   • Radius: $r = \sqrt{36} = 6$

   Example 2: Determining if a Point Lies on the Circle

   To determine whether a point $(x_1, y_1)$ lies inside, on, or outside the circle $(x - h)^2 + (y - k)^2 = r^2$, evaluate the value $S = (x_1 - h)^2 + (y_1 - k)^2$:

   • If $S = r^2$, the point lies on the circle.
   • If $S < r^2$, the point lies inside the circle.
   • If $S > r^2$, the point lies outside the circle.