This time I have an "I don't understand a concept" question.
So right now I'm doing quadratics, more specifically graphing circles.
I got a question where they asked me to find the center and radius of the circle from an equation (attached).
I decided to see if I could find any x intercepts first, so I used the old method for finding x-intercepts we learned in the section on parabolas.
I make y=0, then solved for x using the null factor theorum (as can be seen).
I got two x values, deduced the center point and radius, and thought I had it worked out.
Looking at the worked solution (attached) I can see they completed the square for y to put the equation in (circle form? )
My question is: Why does my method not work? It seems to follow the rules I've been taught previously, but it produces a different result than the complete the square method.
Much appreciated,
Corey
Not sure if this is a misconception of yours or not, but I would first like to address this regardless: a circle is not a quadratic. Although there is a degree two x variable, there's also a degree two y variable in the general equation which prevents us from classifying a circle as a quadratic. You may then ask: what type of equation is a circle? Well, for the purposes of the syllabus, a circle is just a circle. That's all there is to it
(If you're interested, you could search up "conics" which can give you more insight into a circle, but that's beyond the Methods curriculum, so don't stress if you don't understand it all). Also apologies if you already know that a circle is not a quadratic.
What you did was find the midpoint of the two x-intercepts, (4,0) and (-4,0), of the circle, getting (0,0) and assuming that this is the center of the circle. The issue with doing this is that, using your x-intercepts, you can only deduce the
x-coordinate of the center, which is 0, in this case. However, we cannot assume that the center has the same y-coordinate as these x-intercepts (i.e. we cannot assume the y-coordinate is zero). Using your method, we would actually have to find the midpoint of the
y-intercepts, which are (0,8) and (0,-2), and take the y-coordinate (3) of the midpoint (which is (0,3) ) to get the y-coordinate of the circle's center. We would then put these pieces of information together to get the center of (0,3).
From this, you can see that the method is a bit lengthy and requires a bunch of substitution and quadratic solving. Not to mention that calculating the radius could also be a disaster (not here, but in cases where neither the x-coordinate nor the y-coordinate of the center are 0). What's worse is that this method cannot be used with all circles, particularly those which have less than 2 x-intercepts and/or less than 2 y-intercepts. For example, try your method with a circle of this equation:
x
2-4x+y
2-4y+7=0
If you tried substituting x=0 or y=0 into the equation, you'll just end up getting a quadratic which cannot be solved. That's why the "complete the square" method is used with circles; you can complete the square with any quadratic you have, and deduce information from there. Ok, I just said that circles are not quadratics. However, you can treat the degree 2 x and y variables as quadratics. The thing is, you complete the square separately for your x and y variables. This process may be a bit confusing at times. In the above equation, you could write -4(x+y). Don't do that, that's keeping the x and y variables in the same brackets and will just lead to a load of confusion.