2 Answers

Joost · Answer 1 · 2017-12-31T09:34:42+0000

we know that

the standard form of the equation of the circle is

(x-h)^(2) +(y-k)^(2)=r^(2)

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

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

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(x^(2)+8x) +(y^(2)-6y)=-21

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

(x^(2)+8x+16) +(y^(2)-6y+9)=-21+16+9

(x^(2)+8x+16) +(y^(2)-6y+9)=4

Rewrite as perfect squares

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

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

the center of the circle is
(-4,3)

the radius of the circle is
$2\ units$

therefore

the answer is

the radius of the circle is
$2\ units$

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