পুনরা‌লোচনা: বৃ‌ত্তের সমীকরণ

The general equation for a circle is $$(x-h)^2+(y-k)^2=R^2$$

Where the center of the circle is at $(h,k)$, and the circle is all the points $(x,y)$ that are $r$ units away from the center $(h,k)$.

The general equation for a circle is $$(x-a)^2+(y-b)^2=r^2$$ Where the center of the circle is at $(a,b)$, and the circle is all the points $(x,y)$ that are $r$ units away from the center $(a,b)$.

Let's figure out how to derive this!

Let $(x,y)$ be a point in the coordinate plane.

This is the distance formula, also known as the Pythagorean Theorem!

$\sqrt{x^2+y^2} = 1$

So this means that the set of all points that are one unit away from the origin is given by $x^2+y^2=1.$

Great! So we've found the equation of a specific circle (this circle is called the unit circle).

Let's say we want to find the equation of a circle in general now. That is, we want a circle whose center is $(a,b)$ and whose radius is $R.$

Great job! You've found the general form for the equation of circle.

Let us assume a problem asks for a graph of a circle, the equation of which is:

$$x^2+ 4x +y^2 +20 y +4=0$$

The most important aspects of the graph that must be found are the radius and the center of the circle.

We begin by combining like terms, which gives us the following:

$$(x^2+ 4x +4) + y^2 +20y =0$$

We then have to complete the square for the $y^2 + 20y$ which leads us to the next step:

$$(x^2+ 4x +4) + (y^2 +20y +100) =100$$

Note that since I added $100$ to the left side to complete the square, I have to add $100$ to the other side to maintain equality.

Finally we can represent the same expression in a form that allows for us to easily find the center and radius:

$$(x+2)^2 + (y+10)^2 =100$$

Remember the general case for the equation of a circle: the center is located at $(h,k)$ and the radius is $r$ in the general formula:

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

When compared with the form we have derived by completing the square, it becomes easy to find that the center is $(-2,-10)$ and the radius is $10$.

And that is how to convert a set of terms into a form from which it is easy to interpret the center and the radius, which makes graphing simpler.

Circles (from the analytic geometry perspective)

The formal definition of a circle is "the set of all points $P$ at a fixed distance (called the radius, $r$) from a given point (called the center, $C$)."

From here, we can use the distance formula: given $C \in \mathbb{R}^2$ and $r \in \mathbb{R}, r > 0$, a circle is the set of all points $P \in \mathbb{R}^2$ such that $||P-C||=r$. We know that $P$ is an unknown, so it has coordinates $(x,y)$. $C$ is fixed, so call its coordinates $(h,k)$. We therefore have $||(x,y)-(h,k)|| = r$, or $||(x-h,y-k)|| = r$, which is, by the definition of the length of a vector, equivalent to $\sqrt{(x-h)^2 + (y-k)^2} = r$, so therefore $(x-h)^2 + (y-k)^2 = r^2$. If you expand this, you get an equivalent equation of the form $x^2 + y^2 + ax + by + c = 0$.

Conversely, if you're given the equation $x^2 + y^2 + ax + by + c = 0$, you can complete the square in $x$ and in $y$ to get $(x + \frac{a}{2})^2 + (y + \frac{b}{2})^2 = \frac{a^2 + b^2}{4} - c$. The center is therefore $(-\frac{a}{2}, -\frac{b}{2})$, and the radius is $\frac{1}{2}\sqrt{a^2+b^2-4c}$ (if the radicand is positive).

In the two-dimensional Cartesian Plane, we can write an equation representing a circle. This equation is the combination of all points on the circle, which happen to connect. The equation is

$(x - a)^2 + (y - b)^2 = r^2$

$(a, b)$ is the center of the circle and $r$ is the radius of the circle. Note that if $a$ and $b$ equal $0$, we get

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

This is the equation for a circle of radius $r$ centered at the origin.

Note that if we know the circle's radius and any two points on the circle, we can find the center of the circle. Similarly, if we know the center and radius of a circle, we can find all points on the circle.

A circle is the locus of points equidistant from a point. Let's call this point $O(x_0,y_0)$ and the distance $r$.

The Pythagorean Theorem states that, given a right triangle with legs $a$ and $b$ and hypotenuse $c$, $a^2+b^2=c^2$. Since the $x$ and $y$ axes are perpendicular, we can apply the Pythagorean Theorem to get $$(x-x_0)^2+(y-y_0)^2=r^2.$$

Here is some extra practice on Equations of Circles, have fun!