Write an equation for the circle

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In summary, we discussed finding the equation for a circle given the endpoints of a diameter, simplifying the square root of 50, and the relationship between rise and run. We also considered the equation for the outermost circle created when a stone is dropped in a lake and the radious increases at a rate of 20cm/s 2.5s after the stone hits the water. The equation for this circle is x^2 + y^2 = 2500.
  • #1
Aya
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Hi, I need help with these questions

Given the endpoints of a diameter, write an equation for the circle.
a) (-3,5) and (1,3)

I tryed this

m= (x1+x2)/2, (y1+y2)/2
= (-3+1)/2 , (5+3)/2
= -2/2, 8/2
= (-1,4)

( x-h)^2 + (y-k)^2=r^2
(x+1)^2 + (y-4)^2=r^2
(1+1)^2 + (3-4)^2=r^2
(2)^2 + (-1)^2=r^2
5=r^2

so, (x+1)^2 + (y-4)^2 = 5

but, this answer in the back of the book is different so can someone tell me what to do?

and

1. how do you simplify root 50
2. is rise y and run x?

***
when a stone is dropped in a lake, circular waves or ripples are created. The enrey point of the stone marks the center of the circles. Suppose the radious increases at the reate of 20cm/s. write anequation for the outermost circle 2.5s after the stone hits the water.

20*2.5=50cm
X^2 + Y^2=50^2
X^2 + Y^2=2500

is that right?
 
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  • #2
That answer looks right to me...What is the answer that is given?

1) To simplify the square root of 50, can you think of a way to write 50 as the product of two numbers, one of which is a perfect square?

2) Yes. You have to be careful though because it really depends on how you set up your axis. There's nothing stopping you from writing a line in the form of x=my+a, in which case the "rise over run" is a bit misleading if you set up your axis as y vertically and x horizontally.
 
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  • #3
^ thanks

1) To simplify the square root of 50, can you think of a way to write 50 as the product of two numbers, one of which is a perfect square?

I thought the two numbers were 25 and ...but how do you rght that? 2root25
simplify es to 2root5, but that's wrong right?

That answer looks right to me...What is the answer that is given?
the answer in the book was x^2+y^2=49
 
  • #4
Are you sure you're checking the answer for the right problem?

You are right that you can write 50 as 25*2

[tex]\sqrt{50}[/tex]
[tex]\sqrt{25\cdot 2}[/tex]

Can you see what to do from there?
 
  • #5
Are you sure you're checking the answer for the right problem?

ya, the book is old maby that answer is wrong.

Can you see what to do from there?
root 25 is 5 soo..

5root2?
 
  • #6
Aya said:
root 25 is 5 soo..

5root2?
Yep, that's correct.
 
  • #7
k thanks for you help
 

1. What is the general equation for a circle?

The general equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

2. How do I write an equation for a circle with a given center and radius?

To write an equation for a circle with a given center (h,k) and radius r, use the equation (x - h)^2 + (y - k)^2 = r^2. Simply substitute the values for h, k, and r into the equation to get the specific equation for the circle.

3. Can I write an equation for a circle with a center that is not at the origin?

Yes, you can write an equation for a circle with a center that is not at the origin. Use the general equation (x - h)^2 + (y - k)^2 = r^2 and substitute the values for h and k to represent the center of the circle.

4. How do I graph a circle using its equation?

To graph a circle using its equation, first determine the center and radius from the equation. Then, plot the center point on the coordinate plane and use the radius to draw the circle. You can plot additional points on the circle by substituting different values for x into the equation and solving for y.

5. Can an equation for a circle have a negative radius?

No, an equation for a circle cannot have a negative radius. The radius of a circle must be a positive value, as it represents the distance from the center to any point on the circle's circumference. If the radius is negative, the equation will not produce a valid circle.

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