05. Distance, Circle and Quadratic Equations

Dated: 30-10-2024

Distance Formula

Notice how there is a triangle?
We can find the distance between Aand B by using Pythagorus thoerem.

\[(hyp)^2 = (perp)^2 + (base)^2\]

\[hyp = \sqrt{(perp)^2 + (base)^2}\]

\[distance = \sqrt{(\Delta y)^2 + (\Delta x)^2}\]

\[distance = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\]

Midpoint

The midpoint of \(\overline{AB}\) is defined to be \(average\) of the coordinate values relative to \(axes\).

Circle

A circle is defined to be a set¹ of points which are equidistant from a certain point called the center of the circle.

\[circle := \{(x, y): r = \sqrt{x^2 + y^2}; r \in \mathbb{R^+}\}\]

General Form

\[ax^2 + by^2 + dx + ey + f = 0\]

Example

Find radius and center of the circle \(x^2 + y^2 - 8x + 2y + 8 = 0\)

\[x^2 + y^2 - 8x + 2y + 8 = 0\]

\[(x^2 - 8x) + (y^2 + 2y) = -8\]

\[(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1\]

\[(x - 4)^2 + (y + 1)^2 = 9\]

\[(x - 4)^2 + (y - (-1))^2 = 3^2\]

\[\because d^2 = r^2 = (x - x_0)^2 + (y - y_0)^2\]

Where \(P(x, y)\) will be any arbitrary point and \(P_0(x_0, y_0)\) is the center.
So the radius is 3 and the center is (4, -1)

Parabola

This parabola can be represented by equation:

\[y = (x - 2) \cdot (x - 5)\]

Notice how \(y\) becomes 0 if either of the terms become 0 and for that, we need x to be equal to either 2 or 5.
Let us try to simplify it

\[y = x \cdot (x - 5) - 2 \cdot (x - 5)\]

\[y = x^2 - 5x - 2x + 2 \cdot 5\]

\[y = x^2 - x(5 + 2) + 2 \cdot 5\]

\[y = x^2 - Sx + P\]

Where S is sum of the roots and P is product of the roots.

General Equation

\[y = ax^2 + bx + c\]

Notice the resemblance?
For the roots, equation takes form of

\[0 = x^2 + \frac{b}{a} \cdot x + \frac{c}{a}\]

Where \(-\frac{b}{a} = sum\) and \(\frac{c}{a} = product\)

Vertex

The tip of the parabola is called its vertex and it exists in middle of the roots

\[middle = \frac{sum}{2}\]

The x-coordinate will be

\[x = - \frac{b}{2a}\]

And we can get the y-coordinate just by putting this in \(y = f(x)\)