To solve we use the following expression:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Here (h, k) is the center of the circle, now we solve:
5. We calculate it as follows:
[tex](6-3)^2+(5-k)^2=13[/tex]
We have to solve now for k:
[tex]\Rightarrow9+(k^2-10k+25)=13\Rightarrow k^2-10k+21=0[/tex]
Now using the quadratic formula we solve:
[tex]k=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(1)(21)}}{2(1)}\Rightarrow\begin{cases}k=7 \\ k=3\end{cases}[/tex]
Now, we have to test each value of k as follows:
We replace the values of k and obtain the following equations:
[tex]\begin{cases}(x-3)^2+(y-7)^2=13 \\ (x+3)^2+(y-3)^2=13\end{cases}[/tex]
Now, we test the point given (6, 5):
[tex]\begin{cases}(6-3)^2+(5-7)^2=13 \\ (6-3)^2+(5-3)^2=13\end{cases}\Rightarrow\begin{cases}13=13 \\ 13=13\end{cases}[/tex]
So, we will proceed as follows since both values are equivalent:
So, both equations represent a circle with radius sqrt(13), with a center at x = 3 and that passes the point (6, 5).
7. We calculate it as follows:
[tex](0-h)^2+(3-0)^2=25[/tex]
Now, we solve for one of the possible values:
[tex]\Rightarrow h^2=16\Rightarrow h=\pm4[/tex]
So, its center is located in either (4, 0= 0r (-4, 0), now we proceed as follows:
[tex]\begin{cases}(x-4)^2+(y-0)^2=25 \\ (x+4)^2+(y+0)^2=25\end{cases}[/tex]
Now, we test the values: given the point (0, 3):
[tex]\begin{cases}(0-4)^2+(3)^2=25 \\ (0+4)^2+(3)^2=25\end{cases}\Rightarrow\begin{cases}25=25 \\ 25=25\end{cases}[/tex]
And we can also see the following graph:
This means, that any of the two equations describe the circle in the text.
