Locus Of Points: Distance From A Fixed Point

Welcome to our exploration of a fascinating concept in mathematics: the locus of points. Today, we're diving into a specific problem where we need to find the locus of point P such that the distance from point P to A(4,3) is 2. This might sound a bit technical at first, but trust me, it's quite intuitive once we break it down. At its core, a locus of points is simply a set of all points that satisfy a particular condition. Think of it as drawing a path or a shape where every single point on that path or shape has a specific characteristic. In our case, that characteristic is its distance from a fixed point.

Let's set the stage. We have a fixed point, which we'll call $A$, and its coordinates are given as $(4,3)$. We are looking for a set of all points, let's call any generic point in this set $P$, such that the distance between $P$ and $A$ is exactly 2 units. Imagine you're standing at point $A$. Now, you want to find all the places you can go where you are precisely 2 steps away from where you are standing. What kind of shape would that trace out? It's a fundamental geometric idea, and understanding it unlocks many other mathematical concepts, from circles to parabolas and beyond. The beauty of this is that it doesn't just apply to 2D space; the concept of a locus extends to 3D and even higher dimensions!

To formalize this, let the coordinates of our moving point $P$ be $(x,y)$. We are given the fixed point $A$ with coordinates $(4,3)$. The condition we need to satisfy is that the distance between $P(x,y)$ and $A(4,3)$ is always equal to 2. How do we express the distance between two points in a Cartesian coordinate system? We use the distance formula, which is derived directly from the Pythagorean theorem. If we have two points $(x_1, y_1)$ and $(x_2, y_2)$, the distance $d$ between them is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Applying this formula to our specific problem, where $P(x,y)$ is our moving point and $A(4,3)$ is our fixed point, and the distance $d$ is 2, we get: $2 = \sqrt{(x - 4)^2 + (y - 3)^2}$. This equation is the mathematical representation of the condition we are given. However, it's often more convenient to work with an equation that doesn't involve a square root. To eliminate the square root, we can square both sides of the equation. So, we have $2^2 = (x - 4)^2 + (y - 3)^2$. This simplifies to $4 = (x - 4)^2 + (y - 3)^2$.

Now, let's pause and look at this equation: $(x - 4)^2 + (y - 3)^2 = 4$. Does this form look familiar? If you've studied conic sections or basic coordinate geometry, this equation should ring a bell. It is the standard form of the equation of a circle! A circle is defined as the set of all points in a plane that are at a fixed distance (the radius) from a fixed point (the center). In our equation, the center of the circle is $(4,3)$, which is precisely our point $A$, and the radius is $\sqrt{4} = 2$, which is the given distance.

Therefore, the locus of point $P$ such that the distance from point $P$ to $A(4,3)$ is 2 is a circle with center $A(4,3)$ and a radius of 2. This geometric interpretation makes the concept very tangible. Imagine drawing a point on a piece of paper at (4,3). Then, take a compass, set its width to 2 units, place the needle at (4,3), and draw a circle. Every single point on that circumference is exactly 2 units away from the center (4,3). That is the locus we've found.

Understanding the Geometric Interpretation

Let's delve deeper into the geometric interpretation of our problem: finding the locus of point P such that the distance from point P to A(4,3) is 2. When we talk about the