Answer: 3
Step-by-step explanation:
Given
The equation of the circle is [tex]x^2+y^2-4x-6y-k=0[/tex]
Using completing the square method
[tex]\Rightarrow x^2+y^2-4x-6y+4-4+9-9-k=0\\\Rightarrow \left( x^2-4x+4\right)+\left( y^2-6y+9\right)-\left( k+4+9\right)=0\\\Rightarrow \left( x-2\right)^2+\left( y-3\right)^2=\left( k+13\right)[/tex]
on comparing with standard form, [tex]r^2=k+13[/tex]
Insert the value of r
[tex]\Rightarrow 4^2=k+13\\\Rightarrow k=16-13\\\Rightarrow k=3[/tex]
