# Thread: Aaaah confused on analytic geometry questions!!

1. ## Aaaah confused on analytic geometry questions!!

Urgent =( I managed to get the other twenty two but I need help on these three!! If anybody is willing to give me an answer and explanation I would be very very grateful!

[size=3]16. Which point on the circle x^2 + y^2 -12x - 4y= 50 is the closest to the origin? Which point is farthest? Explain.

2. 1. 3 diameters + 1 circumference

2. 4 diameters + 1 circumference

(for 1 and 2 ... see why?)

3. complete the square to get the circle equation in the form

$(x-h)^2 + (y-k)^2 = r^2$

then sketch the circle to see what point is closest/farthest to the origin

3. Originally Posted by skeeter
1. 3 diameters + 1 circumference

2. 4 diameters + 1 circumference

(for 1 and 2 ... see why?)

3. complete the square to get the circle equation in the form

$(x-h)^2 + (y-k)^2 = r^2$

then sketch the circle to see what point is closest/farthest to the origin
Thanks you so much! I get it now, thanks SO MUCH.

I'm not sure if I'm supposed to post an entire new thread so I'll post this here, I want to see if I'm off or not.

The question is:

An isosceles triangle has two sides of length p and one of length m. IN terms of these lengths, write calculator-ready formulas for the sizes of the angles of this triangle.

For one of the two base angles, would the formula be something along the lines of:

inverse cos= .5m/p

Is that right or totally off??

4. I put the three circles on a set of coordinate axes.

$(x-6)^{2} + y^{2} = 6^{2}$

$(x+6)^{2} + y^{2} = 6^{2}$

$x^{2} + (y-6\sqrt{3})^{2} = 6^{2}$

No notice a few things.

If you connect the centers, you get an equilateral triangle. Draw this triangle. Your band obviously is longer than this.

What do you suppose you get if you extend the three tangents until they intersect? Draw this triangle. Your band obviously is shorter than this.

Draw all six radii, two from each center, intersecting the tangents on the circle. These radii are perpendicular to the tangents.

As far as the four pipes, is there a unique solution? You'll have to explore that.