Give the equation of the circle centered at the origin and passing through the point (-3,0)

Answer 1

The equation of a circle is:

(x-h)² + (y-k)² = r²

where (h,k) is the location of the center and r is the radius. If the circle is cnetered at the origin, (0,0), then its equation is:

(x-h)² + (y-k)² = r²

(x-0)² + (y-0)² = r²

x² - y² = r²

Now we just need to find the value of r, the radius. The point (-3,0) is on the circle so the distance from the center (0,0) to (-3,0) is equal to the radius. Can you finish it?

Answer 2

The equation of a circle is in the form (x-h)² + (y-k)² = r², where (h,k) is the center of the circle, (0,0), and "r" is the radius. Because the point (-3,0) is on the x-axis the radius is 3. Putting these numbers into our formula we get:

(x-h)² + (y-k)² = r² equation of a circle

(x-0)² + (y-0)² = 3² substituting for h, k and r

x² + y² = 9 result

Hope this helps!

Answer 3

The origin is the point (0,0) where both axes meet. If the circle is centered at (0,0) and passes through a point 3 units from its center, then we know the radius of the circle is 3. The equation of any circle is given by:

(x-a)² + (y-b)² = r² Where r is the radius of the circle, and the point (a,b) is the center of the circle

In this case, the problem is greatly simplified by the circle being centered at the origin, which means:

(a,b) = (0,0)

If we plug in all the information we've been given the equation is:

(x-0)² + (y-0)² = 3²

All you need to do from here is simplify, and solve for y if you need to.

Hope this helps!