Video: Finding the Center and Radius of a Circle Using the Equation of a Circle

Find the centre and radius of the circle π‘₯Β² + 𝑦² + 10π‘₯ βˆ’ 18𝑦 + 42 = 0.

03:43

Video Transcript

Find the centre and radius of the circle π‘₯ squared plus 𝑦 squared plus 10π‘₯ minus 18𝑦 plus 42 equals zero.

So we’ve been given the equation of a circle in its expanded form, and we’re asked to find its centre and radius. In order to do so, we need to rearrange this equation so that we have it in the standard form for the equation of a circle, the centre-radius form. This standard form is π‘₯ minus β„Ž squared plus 𝑦 minus π‘˜ squared equals π‘Ÿ squared, where the centre of the circle is the point with coordinates β„Ž, π‘˜ and the radius of the circle is π‘Ÿ.

In order to rearrange the equation of the circle into this form, we need to complete the square. First, I’m going to reorder the terms so that the π‘₯ squared and the π‘₯ terms are together, the 𝑦 squared and the 𝑦 terms are together, and then I’ve subtracted 42 from both sides. So I now have negative 42 on the right-hand side of the equation.

So I am going to complete the square in π‘₯ first of all. I need to halve the coefficient of π‘₯, so that’s 10 and square this value. So half of 10 is five, and when I square this, I get 25. However, adding this 25 in, I have unbalanced my equation. So I need to add it onto the right-hand side as well.

Now we complete the square in 𝑦. So we look at the coefficient of 𝑦 which is negative 18. Halving that, we have negative nine and then squaring that, we have 81. Again, if I only add this to one side of the equation, then I’ve unbalanced it. So I also need to add 81 to the right-hand side.

Now looking at the first three terms, we can see that we have a perfect square. π‘₯ squared plus 10π‘₯ plus 25 is equivalent to π‘₯ plus five all squared. And that’s not a coincidence, that’s by design because we’ve been using the completing the square method. So we replace the first three terms with π‘₯ plus five all squared.

Now let’s look at the next lot of terms, those in 𝑦. We have 𝑦 squared minus 18𝑦 plus 81, which is also a perfect square. It’s equal to 𝑦 minus nine all squared. So we replace these next two terms with 𝑦 minus nine all squared. On the right-hand side, we have negative 42 plus 25 plus 81 which simplifies to 64.

Now if we compare the equation of our circle with the standard form, we can see that we’re nearly there. The value of π‘˜, in particular, we can see is nine. If we look at the π‘₯ part, then we have π‘₯ plus five all squared, and we’re comparing this with π‘₯ minus β„Ž all squared. This means that we must be subtracting negative five in order to reach π‘₯ plus five. So we have π‘₯ minus negative five all squared. And now we can see the value of β„Ž as well.

Finally, let’s look at the value of π‘Ÿ. On the right-hand side of the equation, we have 64. And 64 is equal to β„Ž squared. So now if we compare this with the standard form, we can see that π‘Ÿ is equal to eight.

So let’s put all this together then and write down the centre and the radius of the circle. We’ve found that the centre of the circle is the point with coordinates negative five, nine and the radius of the circle is eight length units.

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