Question

Using the following equation, find the center and radius of the circle.

Answer 1

Answer:

Center: $(-2,3)$

Radius = 3.

Step-by-step explanation:

We have been given an equation $x^2+4x+y^2-6y=-4$. We are asked to find the center and radius of the circle.

To find the center and radius of the circle, we need to convert our given equation in standard form of circle $(x-h)^2+(y-k)^2=r^2$, where, (h,k) is center of circle and r represents radius of circle.

To complete the square, we need to add $(\frac{b}{2})^2$ to both sides of our given equation.

$(\frac{4}{2})^2=2^2=4$

$(\frac{6}{2})^2=3^2=9$

$x^2+4x+4+y^2-6y+9=-4+4+9$

$x^2+4x+4+y^2-6y+9=9$

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

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

Therefore, the point $(-2,3)$ is the center of circle and radius of circle is 3 units.

Answer 2

(x²+4x+4)-4+(y²-6y+9)-9=-4 (x+2)²+(y-3)²=9 (x+2)²+(y-3)²=3² 
Circles follow the form: (x-a)+(y-b)=r²
Therefore the center of the circle is at (-2,3)
and the radius is 3.