Checking: Circle Equations Questions

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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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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.
 
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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