Animated Solution for Mathematics - Circles: The point diametrically opposite to the point $P(1,0)$ on the circle $x^2+y^2+2x+4y-3=0$ is

Welcome, future engineer! Today, we are going to explore the beautiful, symmetrical world of coordinate geometry.

We have a circle defined by the equation $x^2+y^2+2x+4y-3=0$ and a point $P(1,0)$ resting on its edge. Our mission is to find the point $Q$ that sits diametrically opposite to $P$.

Imagine standing at $P$ and walking in a straight line through the very heart of the circle to reach the other side. That destination is our goal.

Before we can find the opposite point, we must understand the circle itself. The general equation of a circle is given by $x^2+y^2+2gx+2fy+c=0$.

By comparing this to our given equation, $x^2+y^2+2x+4y-3=0$, we can extract the vital parameters. We see that $2g=2$, which means $g=1$, and $2f=4$, which means $f=2$.

The center of any circle in this form is located at $C(-g,-f)$. Therefore, our center $C$ is at $(-1,-2)$. This center is the anchor of our entire problem.

Now, let's visualize the geometry. A diameter is a line segment that passes through the center $C$ and connects two points on the circle, $P$ and $Q$.

Because the distance from the center to any point on the circle is the radius, the center $C$ must be exactly halfway between $P$ and $Q$. This is the geometric soul of the problem: $C$ is the midpoint of the segment $PQ$.

We can now use the midpoint formula, which states that the coordinates of the midpoint are the average of the coordinates of the endpoints:

We know $P(1,0)$ and $C(-1,-2)$. Let the unknown point $Q$ be $(x,y)$. Substituting these into our midpoint formula, we get:

This gives us two simple, independent linear equations. For the x-coordinate, we have:

For the y-coordinate, we have:

The Beauty of Symmetry

And there we have it! The point diametrically opposite to $P(1,0)$ is $Q(-3,-4)$.

It is a perfect, logical conclusion. By understanding the relationship between the center and the diameter, we turned a potentially complex geometric problem into a simple algebraic exercise.

Keep this mindset—always look for the symmetry, always find the center, and the math will reveal itself to you. You are doing great!