<body><script type="text/javascript"> function setAttributeOnload(object, attribute, val) { if(window.addEventListener) { window.addEventListener('load', function(){ object[attribute] = val; }, false); } else { window.attachEvent('onload', function(){ object[attribute] = val; }); } } </script> <div id="navbar-iframe-container"></div> <script type="text/javascript" src="https://apis.google.com/js/plusone.js"></script> <script type="text/javascript"> gapi.load("gapi.iframes:gapi.iframes.style.bubble", function() { if (gapi.iframes && gapi.iframes.getContext) { gapi.iframes.getContext().openChild({ url: 'https://www.blogger.com/navbar.g?targetBlogID\x3d14084555\x26blogName\x3dPre-Cal+30S\x26publishMode\x3dPUBLISH_MODE_BLOGSPOT\x26navbarType\x3dBLUE\x26layoutType\x3dCLASSIC\x26searchRoot\x3dhttp://pc30s.blogspot.com/search\x26blogLocale\x3den_US\x26v\x3d2\x26homepageUrl\x3dhttp://pc30s.blogspot.com/\x26vt\x3d931551856370134750', where: document.getElementById("navbar-iframe-container"), id: "navbar-iframe" }); } }); </script>

Tuesday, October 25, 2005

Scribing For Circles

Hi guys! Today we started a new unit on equations of circles. The first thing we did was talk about the distance formula. You might remember this from last year. This equation is used to find the distance from one point to another.
The next thing we did was use this piece of information to find the radius of a circle. If the coordinates (h,k) is the center point of the circle and (x,y) is one of the endpoints then the distance from both points is the radius.


To get rid of the square root sign you have to square both sides. You'll end up with an equation like this. This is the equation of a circle.


Two things you need to know about a circle is its radius and its center point. From this equation, you can retrieve (h,k). Remember that the actual coordinates of (h,k) is the negative inverse or opposite from h and k in the equation. For the radius, remember to square root it for your answer because in the equation, it's not r.

ex.1: (x-3)²+(y+2)²=64

center=(3,-2) (negative inverses of h and k)

radius=8 (square root of 64)

ex.2: x²+(y+5)²=20

center=(0,-5) (opposite of h and k)

radius=2 root 5 (square root 20 then simplify)

The next thing we learned was how to write an equation if we were given the information.

ex. center=(-2,8)

radius=

(x+2)²+(y-8)²=49/4

To retrieve this answer the first thing you do is find the opposites of (-2,8) which is 2 and -8 then plug that in to the equation. To get the radius, change 3½ to an improper fraction, 7/2. The next step is to square it and you'll get an answer of 49/4. Leave it as a fraction. Remember, fractions are our friends.

When you have an equation like x²+4x+y²-2y=4, in order to find the center and radius, you must complete the square.

ex. x²+4x+y²-2y=4

(x²+4x+4)+(y²-2y+1)=4+4-1

(x+2)²+(y-1)²=9

center=(2,-1)

radius=3

After, we were given plenty of time to do exercise 21.

Well, that's all that happened today. Good luck on your test tomorrow. The next scribe is Pamela.

2 Comments:

At 10/25/2005 8:54 PM, Blogger Richy said...

Hahaha. Diagrams in 3D. Hahaha. Crrrazy!!! Excellent Work.

 
At 10/25/2005 9:23 PM, Blogger AichelleS. said...

three words...A-MAZ-ING hahaha I love the colours and graphics

 

Post a Comment

Links to this post:

Create a Link

<< Home