Question

What is the radius of a circle whose equation is x2 y2 8x – 6y 21 = 0?

Answer 1

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 $2\ units$

Answer 2

Answer:

Should be (A) on ED

Step-by-step explanation: