Answer:
The question can be explained using the image below
The given coordinates are
![\begin{gathered} A\left(-4,6\right)\Rightarrow\left(x_1,y_1\right) \\ C\left(3,5\right)\Rightarrow\left(x,y\right) \\ B\left(x_2,y_2\right)\Rightarrow \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r6qk4gbpujvmw2q6h7koqb6n3ie8xtdfdz.png)
The center is the midpoint of Diameter AB, therefore, we will use the formula for the midpoint of a line given below as
![\left(x,y\right)\Rightarrow(\left(x_1+x_2\right))/(2),(\left(y_1+y_2\right))/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/3kmu51kn91uvsiwph6atjw6awa7phn75rv.png)
By substituting the values, we will have
![\begin{gathered} (x,y)\operatorname{\Rightarrow}((x_(1)+x_(2)))/(2), ((y_(1)+y_(2)))/(2) \\ \left(3,5\right)\Rightarrow(\left(-4+x_2\right))/(2),(\left(6+y_2\right))/(2) \end{gathered}]()
By equating both equations, we will have
![\begin{gathered} ((-4+x_(2)))/(2)=3 \\ -4+x_2=2*3 \\ -4+x_2=6 \\ x_2=6+4 \\ x_2=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hnoq6dloh2rgfz50cs62bejv1sfms19qis.png)
![\begin{gathered} ((6+y_(2)))/(2)=5 \\ 6+y_2=2*5 \\ 6+y_2=10 \\ y_2=10-6 \\ y_2=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ulp0l8gmvtrnrtr5b6lxzotiix9cwjscai.png)
Hence,
The coordinate of B is
![\Rightarrow\left(10,4\right)](https://img.qammunity.org/2023/formulas/mathematics/college/3rlbb7o1s5genicrvq16xn29cw8eqpruu6.png)