[tex] x^2 + y^2 - 6x + 8y + 21=0\\
x^2-6x+9+y^2+8y+16-4=0\\
(x-3)^2+(y+4)^2=4\\\\
\text{center}=(3,-4)\\
r=\sqrt 4=2 [/tex]
Answer: The center of the circle is (3, -4) and radius is 2 units.
Step-by-step explanation: We are given to find the center and radius of the circle defined by the following equation :
[tex]x^2+y^2-6x+8y+21=0~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
The standard equation of a circle with center (h, k) and radius r units is given by
[tex](x-h)^2+(y-k)^2=r^2.[/tex]
From equation (i), we have
[tex]x^2+y^2-6x+8y+21=0\\\\\Rightarrow (x^2-6x+9)+(y^2+8y+16)-9-16+21=0\\\\\Rightarrow (x-3)^2+(y+4)^2-4=0\\\\\Rightarrow (x-3)^2+\{y-(-4)\}^2=2^2.[/tex]
Comparing the above equation with the standard equation of a circle, we get
center, (h, k) = (3, -4) and radius , r = 2 units.
Thus, the center of the circle is (3, -4) and radius is 2 units.
