How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^2 = 25$, I got that $(x + 2)^2 = 0$, so $x = -2$, so $y = -\frac{3}{4} \cdot (-2) - \frac{13}{2} = -5$.
How to prove that this is the only solution?
 A: As the other answers said, your argument already proved there is only one solution.
You can make an independent check by solving the equations graphically.
$(x−1)^2 +(y+1)^2 =25$ is the equation of a circle, radius $5$, center $(1, -1)$. The second equation is another circle, and the circles have two intersection points. The third equation represents a straight line, which goes through just one of the intersection points.
A: I assume that you have checked that $(x,y) = (-2,-5)$ is indeed a solution.
Why is it the only one?
It is, because you have shown that your three equations must imply that $(x+2)^2=0$, which in turn implies $x=-2$. In other words, whatever solutions $(x,y)$ the three original equations have, you have shown that they must all have the property that $x=-2$.
But given that $x=-2$, the last of the three equations show that we must have $y=-5$. In conclusion, if there is a solution at all to your three equations, then it must be $(x,y)=(-2,-5)$ (and you do need to check that this is a solution, because you can easily end up in the case where there are no solutions at all).
A: $$
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
(x - 1)^2 + \left(\left(-\frac{3}{4}x - \frac{13}{2}\right) + 1\right)^2 = 25 \\
(x + 5)^2 + \left(\left(-\frac{3}{4}x - \frac{13}{2}\right) + 9\right)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\frac{25}{16}\left(x^2+4x+20\right) = 25 \\
\frac{25}{16}\left(x^2+4x+20\right) = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x^2+4x+20 = 16 \\
x^2+4x+20 = 16 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x^2+4x+4 = 0 \\
x^2+4x+4 = 0 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
(x+2)^2 = 0 \\
(x+2)^2 = 0 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
x=-2 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
x=-2 \\
y = -\frac{3}{4}(-2) - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
x=-2 \\
y = -5
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
y = -5
\end{cases}
$$
A: The first two are equations of circles. Two circles may have no point in common, or only one tangency point, or two points of intersection or all points if they're actually the same circle.
It's easy to check this is not the first nor the last case. Now, subtract the first equation from the second one:
$
\begin{cases}
(x^2 - 2x + 1) + (y^2 + 2y + 1) = 25 \\
(x^2 + 10x + 25) + (y^2 + 18y + 81) = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
$
\begin{cases}
x^2 - 2x + 1 + y^2 + 2y + 1 = 25 \\
12x + 24 + 16y + 80 = 0 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
Now the second and third equation make a system of linear equations, so they may have only zero, one or infinity of satisfying points – but certainly not two. Q.E.D.
