Video: Finding the General Form of the Equation of a Circle

Give the general form of the equation of the circle center (8, βˆ’2) and diameter 10.


Video Transcript

Give the general form of the equation of the circle center eight, negative two and diameter 10.

So we have two key pieces of information about this circle. We know its center and its diameter, and we’re asked to find its equation. Let’s recall the center radius form of the equation of a circle, which tells us that if the circle has center with coordinates β„Ž, π‘˜ and a radius π‘Ÿ, then its equation is given by π‘₯ minus β„Ž squared plus 𝑦 minus π‘˜ squared is equal to π‘Ÿ squared.

The information we’ve been given nearly aligns with this. We can see the values of β„Ž and π‘˜ for the center of the circle are eight and negative two, but we need to be careful because we have been given the diameter of the circle, not the radius.

In order to find the value of π‘Ÿ, we need to halve the diameter. This is really important, and a very common mistake would be to use the value of 10 as the radius of the circle here, which would of course be incorrect. We need to halve it to get that the value of π‘Ÿ is five.

So now let’s substitute the values of β„Ž, π‘˜, and π‘Ÿ into the equation of our circle. Substituting eight for β„Ž and negative two for π‘˜ and five for π‘Ÿ, we now have π‘₯ minus eight all squared plus 𝑦 minus negative two all squared is equal to five squared.

Now firstly, we’ll just neaten this up a little bit. 𝑦 minus negative two is better written as 𝑦 plus two, and five squared is 25, so we have π‘₯ minus eight all squared plus 𝑦 plus two all squared is equal to 25. Now this is the equation of our circle in center radius form.

But looking at the question, it asked us to give our answer in the general form. This just means that we need to perform a bit of algebraic manipulation on our answer. So what we’re going to do is we’re going to expand each of these brackets, and then we’re going to simplify the result.

So I’ve just written these square brackets out as a bracket multiplied by itself, π‘₯ minus eight multiplied by π‘₯ minus eight and 𝑦 plus two multiplied by 𝑦 plus two, because a very common mistake when squaring a bracket, for example, the first one, is to think that the answer is π‘₯ squared plus 64 or perhaps π‘₯ squared minus 64, neither of which is correct.

If you think about it in terms of multiplying a pair of brackets out, then hopefully that will jog your memory that there will be π‘₯ terms as well. So when I expand that first pair of brackets, I get π‘₯ squared minus eight π‘₯ minus eight π‘₯ plus 64, not just π‘₯ squared plus or minus 64.

Expanding the second pair of brackets, I have 𝑦 squared plus two 𝑦 plus two 𝑦 plus four, and then this is all equal to 25. Now we just need to tidy up the answer a little bit, so we’ll start with π‘₯ squared plus 𝑦 squared.

Now if we look at the π‘₯ terms, we have negative eight π‘₯ minus another eight π‘₯, so we have minus 16π‘₯ overall. Looking at the 𝑦 terms, we have plus two 𝑦 plus two 𝑦, so we have plus four 𝑦.

Now looking at the constant terms, those with no π‘₯s or 𝑦s or π‘₯ squared or 𝑦 squared, I have 64 plus four on the left-hand side, which is 68, but I’d like to bring that 25 over, so I’m going to subtract 25 from both sides of the equation, and that will give me plus 43.

Now because I subtracted that 25, I now have zero on the right-hand side of the equation. So now we have the equation of our circle in the general expanded form: π‘₯ squared plus 𝑦 squared minus 16π‘₯ plus four 𝑦 plus 43 is equal to zero. And the order of the terms there is just the standard conventional order: π‘₯ squared, 𝑦 squared, π‘₯, 𝑦, and then the constant term.

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