What is the radius of a circle whose equation is x2+y2+8x−6y+21=0? 2 units 3 units 4 units 5 units

Given the equation of the circle:

x^2+y^2+8x−6y+21=0

Equation of the circle is represented in the form

x^2 + y^2 = r^2

Solution:

In order to find the radius of the circle, determine first its final equation.

x^2+y&2+8x−6y+21=0

Transpose 21 on the other side of the equation, and supply the missing terms on the x and y variable in order for it to be a perfect square. Note that when transposing the sign changes, and when supplying for the missing term, whatever the number that is added to the left is also added to the right side of the equation.

x^2+y^2+8x−6y=-21

(x^2+8x+ 16)+(y^2−6y+9)=-21 + 16 + 9

Noticed that the variables of x and y can be simplified in its factored form, since they are a perfect square.

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

Basing from the formula x^2 + y^2 = r^2, we know that 4 is still r^2, that is why we need to take the square root of 4 in order to get the value of its radius.

r = 2

The radius of the unit circle given the equation is 2 units.