Checking: Circle Equations Questions

  • Thread starter DizzyDoo
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In summary, the student tried to find an equation for the circle C based on the given information, but failed to solve the equation. They then worked out another equation for the circle based on completing the square. The answers to the equation were (x+2)² - 4 + (y+3)² - 9 - 17 = 0.
  • #1
DizzyDoo
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[SOLVED] Checking: Circle Equations Questions

Last time I got people to check my homework here, I got things marked wrong when they where said to be right. Let's see if it happens again...

Homework Statement



(1) The circle C has centre (3, 4) and passes through the point (8, 8).
Find an equation for C.

(2) Find the centre and radius of the circle x² + y² + 4x + 6y – 17 = 0

(3) A(4, 0) and B(3, 5) are the end points of a diameter of the circle C.
Find (a) the exact length of AB,
(b) the coordinates of the midpoint P of AB,
(c) an equation for the circle C.

Homework Equations



None

The Attempt at a Solution



Right here is my working for (1)

Since we know the circle centre and a point the circle passes through, I can use pythag to work on the radius right? It seems a little too simple to work, but I can't see anything wrong with it. After drawing a clear diagram;

(4+8)² + 5² = r²
r² = 144 + 25
r² = 169
r = 13

Therefore the equation must be (x-3)² + (y-4)² = 13
Correct?

Now, my working for (2);

The equation needs to be factorised, so I have to "complete the square" to get the stuff inside the brackets. This is usually where I go wrong, so check this bit carefully please!

(x+2)² - 4 + (y+3)² - 9 - 17 = 0

Rearraged:

(x+2)² + (y+3)² = 17

Ah, I have to go now, but I'll post the answers to 3 quickly:

a) I got 5.09
b) x = 3.5
y = 2.2
and
c) (x+3.5)² + (y+2.2)² = 5.09

Sorry about that lack of working out. Thank you for your time if you've read this far!
 
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  • #2
1) looks fine. I would make 2) to be (x+2)^2+(y+3)^2=30. Is that just a typo? For 3) they ask for the EXACT value of the length of AB, so I would write sqrt(26) rather than the approximation 5.09.
 
  • #3
Dick said:
1) looks fine. I would make 2) to be (x+2)^2+(y+3)^2=30. Is that just a typo? For 3) they ask for the EXACT value of the length of AB, so I would write sqrt(26) rather than the approximation 5.09.

Thanks! Gotta hand it in later today, so we'll see if I've got them right.
 

Related to Checking: Circle Equations Questions

What is a circle equation?

A circle equation is a mathematical expression that describes the relationship between the x and y coordinates of points on a circle. It is typically written in the form (x-h)^2 + (y-k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.

How do I check if a given equation is a circle equation?

To check if an equation is a circle equation, you can compare it to the standard form (x-h)^2 + (y-k)^2 = r^2. If the equation matches this form, then it is a circle equation. You can also graph the equation and see if it forms a circle.

What are some common mistakes to avoid when checking circle equations?

One common mistake is not squaring the x and y terms when trying to match the standard form. Another mistake is forgetting to subtract the h and k values from the x and y terms in the equation. It is also important to check that the equation has a consistent sign for the squared terms (either both positive or both negative).

Can a circle equation have a negative radius?

No, a circle with a negative radius does not exist. The radius of a circle represents the distance from the center to any point on the circle, so it must be a positive value. If an equation yields a negative radius, it is not a circle equation.

How are circle equations used in real life?

Circle equations are used in many fields, including physics, engineering, and architecture. They can be used to calculate the circumference and area of a circle, which is important for designing circular objects like wheels and gears. They are also used in navigation and mapping to plot the locations of objects on a circular surface, such as the Earth.

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