Answer:
Option (c) is correct.
Center of circle is at (6, -3)
Step-by-step explanation:
Given : The equation of circle is \(x^2+y^2-12x+6y-19=0\)
We have to find the center of given circle and choose from the given options.
The standard equation of circle with center (h,k) and radius r is written as
\((x-a)^2+(y-b)^2=r^2\)
Consider the given equation \(x^2+y^2-12x+6y-19=0\)
Adding 19 both side, we have,
\(x^2-12x+y^2+6y=19\)
Group x and y terms together, we have,
\(\left(x^2-12x\right)+\left(y^2+6y\right)=19\)
Add both side 36 to make x a perfect square term,
\(\left(x^2-12x+36\right)+\left(y^2+6y\right)=19+36\)
Using identity \((a+b)^2=a^2+b^2+2ab\), we have,
\(\left(x-6\right)^2+\left(y^2+6y\right)=19+36\)
Similarly, for y , adding 9 both side,
\(\left(x-6\right)^2+\left(y^2+6y+9\right)=19+36+9\)
\(\left(x-6\right)^2+\left(y+3\right)^2=64\)
rewriting in standard form , we have,
\(\left(x-6\right)^2+\left(y-\left(-3\right)\right)^2=8^2\)
Thus, Center of circle is at (6, -3)
Option (c) is correct.
