Solve for the equation of a circle if the center is origin and contains ( negative eight, six )

Question

center: origin contains: (-8,6)

Answer 1

To start off, here's the standard form for the equation of a circle.

 

(x - h)² + (y - k)² = r²

 

The center is at (h, k), and r is the length of the radius. If the center is at the origin, then that means (h, k) is (0, 0), so plug in h = 0 and k = 0.

 

(x - h)² + (y - k)² = r²

(x - 0)² + (y - 0)² = r²

 

Because subtracting zero doesn't do much, this equation can be simplified a bit to look like this.

 

x² + y² = r²

 

The only thing left to do is to find r². Maybe we can do something with that point at (-8, 6). Plug this in for x and y, so x = -8 and y = 6.

 

(-8)² + (6)² = r²

64 + 36 = 100

 

This means that r² is 100. This was the only unknown term left in the equation, so plug this in now.

 

x² + y² = r²

x² + y² = 100

 

I hope this helps, and good luck with your math class!

Answer 2

Standard equation for a circle:

 

(x-h)² + (y-k)² = r²

 

Where (h,k) = center

and r= radius

 

Since we know the center is at the origin, (0,0)

 

the equation becomes:

 

(x-0)² + (y-0)² = r²

 

x² + y² = r²

 

To ensure the point (-8,6) is contained within the circle, we must find the distance to this pint from the center, origin.

 

You can make a right triangle with this point and the y-axis. The hypotenuse will be the distance to the point, sides 8, and 6.

 

8² + 6⁶ = 64 + 36 = 100

 

so the radius needs to be 10

 

x² + y² = 100

 

 

 

 

Answer 3

The general formula of a circle is (x-h)^2+(y-k)^2=r^2

Where h is your x value from the origin, so -(-8)=8 --> (x+8)^2

k is your y value from the origin, so -(6)=-6 --> (y-6)^2

and r is the radius of your circle