Introduction to Conic Sections and The Circle

Introduction to Conic Sections

A conic section (or simply a conic) is a curve you get when you slice through a double-napped cone with a plane. 🔪 Imagine a cone like an hourglass. Depending on the angle of your slice, you can create four different types of curves: a circle, an ellipse, a parabola, or a hyperbola.

• Circle: Formed when the plane is perpendicular to the cone's axis.
• Ellipse: Formed when the plane is tilted but still cuts across one nap of the cone.
• Parabola: Formed when the plane is parallel to the side of the cone.
• Hyperbola: Formed when the plane is steep enough to cut through both naps of the cone.

The Locus Definition of a Conic

Another way to define a conic is by using a special property involving distances. A conic section is the locus (path) of a point P that moves in a plane so that the ratio of its distance from a fixed point (the focus, S) to its perpendicular distance from a fixed line (the directrix, L) is a constant. This constant ratio is called the eccentricity (e).

$$\frac{\text{Distance from Focus (SP)}}{\text{Distance from Directrix (PM)}} = e \implies SP = e \cdot PM$$

The value of eccentricity tells you exactly which conic you're looking at:

• If e = 0, the conic is a Circle. (The directrix is at infinity).
• If e = 1, the conic is a Parabola. (The point is equidistant from the focus and directrix).
• If 0 < e < 1, the conic is an Ellipse. (The point is closer to the focus than the directrix).
• If e > 1, the conic is a Hyperbola. (The point is farther from the focus than the directrix).

The Circle 🔵

A circle is the set of all points in a plane that are at a fixed distance (the radius, r) from a fixed point (the center, (h,k)).

Equations of a Circle:

• Standard (Center-Radius) Form: This form comes directly from the distance formula. For any point (x,y) on the circle, its distance from the center (h,k) is r.

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

• General Equation: If you expand the standard form, you get the general second-degree equation for a circle. Any equation in this form represents a circle.

  $$x^2+y^2+2gx+2fy+c=0$$ From this, you can quickly find the circle's properties:

  • Center: $(-g, -f)$
  • Radius: $r = \sqrt{g^2+f^2-c}$
  • For a real circle to exist, the term under the square root must be positive, so $g^2+f^2-c > 0$.

• Diameter Form: This is a neat shortcut. If you know the endpoints of a diameter, $(x_1, y_1)$ and $(x_2, y_2)$, any other point $(x,y)$ on the circle forms a 90° angle with the diameter. This geometric fact leads to the equation:

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

• Parametric Equation: This form is useful for calculus and describing motion. It defines the coordinates of any point on the circle in terms of an angle $\theta$.

  $$x = h+r\cos\theta, \quad y = k+r\sin\theta$$

  Here, $\theta$ is the parameter, representing the angle from the center to the point, measured from the horizontal. It ranges from $0 \le \theta < 2\pi$.

Example 1: Finding the Equation from Center and Radius

Question: Find the equation of the circle with center (-3,2) and radius 4.

Solution: This is a direct application of the center-radius form: $(x-h)^2+(y-k)^2=r^2$. Here, $(h,k)=(-3,2)$ and $r=4$.

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

✅ Standard Form: $(x+3)^2 + (y-2)^2 = 16$

Expanding this gives the general form: $x^2+6x+9 + y^2-4y+4 = 16$

✅ General Form: $x^2+y^2+6x-4y-3=0$.

Example 2: Finding Center and Radius from General Form

Question: Find the center and radius of the circle $x^2+y^2-8x+10y-12=0$.

Solution: We compare the given equation with the general form $x^2+y^2+2gx+2fy+c=0$.

• $2g = -8 \implies \mathbf{g = -4}$.
• $2f = 10 \implies \mathbf{f = 5}$.
• $c = -12$.

Now, we use the formulas for center and radius:

• Center: $(-g, -f) = \mathbf{(4, -5)}$.
• Radius: $r = \sqrt{g^2+f^2-c} = \sqrt{(-4)^2+5^2-(-12)} = \sqrt{16+25+12} = \mathbf{\sqrt{53}}$.

Example 3: Using the Diameter Form

Question: Find the equation of the circle whose diameter has endpoints A(-2,3) and B(4,5).

Solution: Using the diameter form $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$ with $(x_1, y_1) = (-2,3)$ and $(x_2, y_2) = (4,5)$.

$$(x-(-2))(x-4) + (y-3)(y-5) = 0$$

$$(x+2)(x-4) + (y-3)(y-5) = 0$$

Now, expand the products: $$(x^2-2x-8) + (y^2-8y+15) = 0$$

Combining terms gives the final equation: $$x^2+y^2-2x-8y+7=0$$

Example 4: Circle Touching an Axis

Question: Find the equation of a circle that is concentric with $x^2+y^2-4x-6y-3=0$ and touches the y-axis.

Solution: Concentric means the circles share the same center. concentric circles

Step 1: Find the center of the given circle. For $x^2+y^2-4x-6y-3=0$, we have $2g=-4 \implies g=-2$ and $2f=-6 \implies f=-3$. The center is $(-g, -f) = \mathbf{(2, 3)}$.

Step 2: Determine the radius of the new circle. The new circle also has its center at (2, 3). For a circle to touch the y-axis, its radius must be equal to the absolute value of the x-coordinate of its center.

So, the radius $r$ is the horizontal distance from the center (2,3) to the y-axis, which is $r = |2| = \mathbf{2}$.

Step 3: Write the equation of the required circle. Using the center-radius form with center (2, 3) and radius 2: $$(x-2)^2+(y-3)^2=2^2$$ $$(x-2)^2+(y-3)^2=4$$