Question

write the equation in standard form for the circle passing through (-3,0) centered at the origin.

Answer 1

Answer:

x² + y² = 9

Step-by-step explanation:

the standard form of a circle equation is

$(x - h)^2 + (y - k)^2 = r^2$

where (h, k) is the center of the circle, and r is the radius.

since the center of this circle is the origin (0, 0), we have the simplified version

$x^2+y^2=r^2$

to get $r^2$ we simply use the given point

$3^2+0^2=r^2$

$9+0=r^2$

$r^2=9$

our equation in standard form is therefore

$x^2+y^2=9$

Answer 2

Answer:

x^2 + y^2 = 9

Step-by-step explanation:

The equation of a circle is (x^2) -a)^2 + (x^2) - b)^2 = r^2, where a and b are the center of the circle. Because the center of the circle is (0,0), it'll just be x^2+y^2 for one half of the equation. If the circle passes through (-3,0), that would mean the radius of the circle is 3. r^2 is 3^2, which is 9, so the equation of the circle is x^2 + y^2=9. Hope this helps!