### Instruction: Let $(x,y)$ be an ordered pair of real numbers that satisfies the equation $x^2+y^2=14x+48y$. What is the minimum value of $x$?### Response: I notice that the equation has $x^2$ and $y^2$ terms, which makes me think of circles.
I wonder if I can rewrite the equation in a form that looks like the equation of a circle.
To do that, I need to complete the square for both $x$ and $y$.
I can rewrite the equation as $(x^2-14x)+(y^2-48y)=0$.
Then, I can add and subtract the appropriate terms to create perfect squares.
For the $x$ part, I need to add and subtract $(14/2)^2=49$.
For the $y$ part, I need to add and subtract $(48/2)^2=576$.
So, the equation becomes $(x^2-14x+49)+(y^2-48y+576)=49+576$.
I can simplify this to $(x-7)^2+(y-24)^2=625$.
Now, I recognize this as the equation of a circle with center $(7,24)$ and radius $25$.
I want to find the minimum value of $x$, which means I want to find the leftmost point on the circle.
That point will have the same $y$-coordinate as the center, which is $24$.
To find the $x$-coordinate, I can plug in $y=24$ into the equation of the circle.
That gives me $(x-7)^2+(24-24)^2=625$.
Simplifying, I get $(x-7)^2=625$.
Taking the square root of both sides, I get $x-7=\pm 25$.
Adding $7$ to both sides, I get $x=7\pm 25$.
Since I want the minimum value of $x$, I choose the negative sign.