Based on the calculations, the radius of this circle with the given equation is equal to 5 units.
Given the following data:
- [tex]x^2+y^2+8x-6y+21=0[/tex]
The equation of a circle.
Mathematically, the standard form of the equation of a circle is given by;
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where:
- h and k represents the coordinates at the center.
- r is the radius of a circle.
Expanding the equation, we have:
[tex]x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0\\\\r=\sqrt{h^2+k^2}[/tex]
Comparing the terms, we have:
[tex]-2hx = 8x\\\\-2h=8\\\\h=\frac{8}{-2}[/tex]
h = -4
[tex]-2ky = -6y\\\\-2k=6\\\\k=\frac{6}{-2}[/tex]
k = 3.
Now, we can determine the radius:
[tex]r=\sqrt{-4^2+3^2}\\\\r=\sqrt{16+9} \\\\r=\sqrt{25}[/tex]
r = 5 units.
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