Question

What is the radius of a circle whose equation is x^2 + y^2 + 8x - 6y + 21 =0

Answer 1

The standard form of a circle is (x-h)²+(y-k)²=r², (h,k) being the coordiantes of the center, r being the radius.

rearrange the equation:
x²+8x+y²-6y=-21
make squares by adding (half of b)². b in this case being 8 for x, and -6 for y:
so add 4² and (-3)²:
x²+8x+4²+y²-6y+(-3)²=-21+4²+(-3)²
(x+4)²+(y-3)²=2²
so the radius is 2

Answer 2

Hello,

x^2+y^2+8x-6y+21=0

Rewritten in the form of a standard circle equation:

(x-(-4))^2+(y-3)^2=2^2

Therefore, the circle properties are:

(a,b)=(-4,3),r=2

The radius of the circle whose equation is x^2 + y^2 + 8x - 6y + 21 =0 is 2.

Faith xoxo