Question

X^2 + Y^2 +4y =3 complete the square on the x and y terms to find the radius and center of a circle

Answer 1

Answer:

Center = (0, -4)

Radius = $\sqrt{7}$ units

Step-by-step explanation:

We are given;

The equation of a circle;

$x^2+y^2+4y=3$

We are required to determine the radius and the center of the circle;

To do this we are going to use completing square method;

$x^2+y^2+4y=3$

Since we don't have the x term then;

$x^2+ox+y^2+4y=3$

We add the square of half the coefficients of x and y on both sides of the equation;

That is;

$x^2+0x+(0^2)+y^2+4y + (2^2)=3+0+(2^2)$

We get;

$x^2+0x+(0^2)+y^2+4y + (2^2)=7$

Taking the squares;

$(x+0)^2+(y+4)^2=7$

But we know when the equation of a circle is written in the form of;

$(x-a)^2+(y-b)^2=r^2$, where (a,b) is the center and r is the radius of the circle.

Thus, in this case;

$(x+0)^2+(y+4)^2=7$

Center = (0, -4)

Radius = $\sqrt{7}$ units