Animated Solution for Mathematics - Circles: If the lines $2x+3y+1=0$ and $3x-y-4=0$ lie along diameter of a circle of circumference $10\pi$, then the equation of the circle is

Imagine you are standing on a vast, empty coordinate plane. You see a circle, perfect and symmetrical.

You are told that two lines, $2x+3y+1=0$ and $3x-y-4=0$, are diameters of this circle.

The beauty of this problem lies in a simple, elegant geometric truth: the intersection of any two diameters of a circle is, by definition, the center of that circle. If you can find where these two lines meet, you have found the heart of the circle.

Solving for the Center

We have our two constraints:

To find the intersection $(x,y)$, we must solve this system simultaneously. Let us use the elimination method.

If we multiply the second equation by $3$, we get $9x-3y=12$. Now, look at what happens when we add this to our first equation:

The $y$ terms, $3y$ and $-3y$, cancel out beautifully, leaving us with $11x=11$. Thus, we find $x=1$.

With $x=1$ in hand, we substitute it back into the second equation: $3(1)-y=4$, which simplifies to $3-y=4$, giving us $y=-1$.

Our center $C(h,k)$ is at $(1,-1)$. We have successfully located the heart of our circle.

Determining the Radius

Now that we have the center, we need the radius $r$. The problem provides the circumference, which is $10\pi$.

Recall the fundamental formula for circumference:

Setting this equal to the given value, we have $2\pi r=10\pi$. Dividing both sides by $2\pi$, we find $r=5$.

We now have the two essential components of our circle: the center $(1,-1)$ and the radius $r=5$.

Building the Equation

With the center $(h,k)=(1,-1)$ and radius $r=5$, we use the standard form of a circle's equation:

Substituting our values, we get $(x-1{)}^2+(y-(-1){)}^2=5^2$, which simplifies to $(x-1{)}^2+(y+1{)}^2=25$.

Expanding the binomials, we get:

Grouping the terms, we have $x^2+y^2-2x+2y+2=25$. Finally, subtracting $25$ from both sides, we arrive at the elegant result:

This matches our target, and we have successfully navigated the geometry and algebra of the circle. Remember, every equation tells a story; you just have to learn how to read it.