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

Asked by maham237 @ in Mathematics viewed by 304 People

Answered by maham237 @


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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