Non-Linear Relationships - The Circle HELP! (but also factorizing equations_

[Luke_8064]

To Whom It May Concern,

I recently just learnt the circle topic of non-linear relationships in Year 10 Mathematics Methods and I am having a hard time to complete the following sets of questions (see "Question type I am stuck with" attachment).

My teacher tried to explain a to me as two sets of unfactorised terms that need to give a perfect square as their solution. So he said that since y² + 8y can be factored into a perfect square but the other can't, you could split the 16 to +4 and -4 so you can perfect square the other term and find the answer.This made sense to me at the time but now when I try to do the rest I get confused because the rest is not like the first, which make sense.

I guess what I am trying to say is... Could someone please explain Q3)b. To me in a way (possibly with visuals) that is somewhat compliant with the rest of the questions and is easier to understand.

Thank you ever so much,
Luke

[fun_jirachi]

Hey there!

What your teacher is essentially trying to tell you is to complete the square such that the equation of the circle is in the form \((x-h)^2 + (y-k)^2 = r^2\), where the centre of the circle is (h, k) and the radius is r. When we expand this, we have that \(x^2-2hx+h^2+y^2-2yk+k^2 = r^2\), and from this, we can surmise what the values of h and k are from the 'linear equation' (even though it's not technically linear, it's more unfactorised, and not even that ). When we complete the square, we ideally add and subtract constants to keep the equation the same, while also allowing us to complete the square. For example, in 3a), we have that:


Notice how we added and subtracted four to complete the square, then moved the four we subtracted to the other side so we were able to manipulate the equation of the circle into the form \((x-h)^2 + (y-k)^2 = r^2\).

For 3b) a similar thing happens:


Basically, it's all about manipulating the equation into a form you can recognise, using methods that keep the equation the same, in this case adding and subtracting the same number. In other cases, you'll need to multiply by a number and its reciprocal or use some identity.

Have a go at the rest yourself! Hope this helps