Video: Finding the Center and Radius of a Circle by Completing the Square

Find the centre and radius of the circle π‘₯Β² + 6π‘₯ + 𝑦² + 18𝑦 + 26 = 0.

04:45

Video Transcript

Find the centre and radius of the circle π‘₯ squared plus six π‘₯ plus 𝑦 squared plus 18𝑦 plus 26 equals zero.

To find the centre and radius of our circle, I’m gonna put it in this form, which is π‘₯ minus π‘Ž all squared plus 𝑦 minus 𝑏 all squared is equal to π‘Ÿ squared, where π‘Ž and 𝑏 are the π‘₯- and 𝑦-coordinates of our centre of our circle and π‘Ÿ is the radius. To be able to enable us to write it in this form, what we’re gonna need to do is we’re gonna need to complete the square.

Just a quick recap of how we complete the square, so we have π‘₯ squared plus π‘Žπ‘₯. Now to complete the square, we’re gonna put it in this form. So we’re gonna have it equal to π‘₯ plus π‘Ž over two all squared minus π‘Ž over two squared. So remembering that we actually halve the coefficient of our π‘₯ with the coefficient of our 𝑦, it’s gonna get it into this form. Right, so let’s apply this to the equation we have on the left-hand side.

So first of all, we will have π‘₯ plus three all squared minus three squared and this is because our coefficient of π‘₯ is six, so we halve it to get the number inside the parenthesis and the number that we subtract after the parenthesis. So that means that six divided by two gives us three.

And then completing the square with 𝑦, we get 𝑦 plus nine all squared minus nine squared again. We get nine because the coefficient of 𝑦 is 18 and half of 18 is nine. So great, we have completed the square for both of those parts. Then we also need to have a plus 26.

Okay, now we can actually simplify our equation, which would give us π‘₯ plus three all squared plus 𝑦 plus nine all squared minus nine minus 81 plus 26. So we get π‘₯ plus three all squared plus 𝑦 plus nine all squared minus 64 equals zero.

And then one final stage to rearrange it so we have it in the form that we’re looking for is actually we’re gonna add 64 to each side. So we’re left with π‘₯ plus three all squared plus 𝑦 plus nine all squared equals 64. So great, we have it in the form π‘₯ minus π‘Ž squared plus 𝑦 minus 𝑏 all squared is equal to π‘Ÿ squared.

Finally, we want to use our equation to find our centre and radius of the circle. Using our equation to find the centre and radius of the circle, we’re required to look back at the circle equation and see how it’s gonna be useful. Well, we’re gonna start by finding the centre of the circle and we’re gonna do this by using our values within the parenthesis. And we can find our centre of circle because when we look back at the original equation, we can see that actually our π‘Ž and 𝑏 values are going to be our π‘₯- and 𝑦-coordinates.

So that means in our circle, we’re gonna have π‘₯-coordinate of negative three and a 𝑦-coordinate of negative nine. You might ask why have you put negative. The reason I have made them negative because actually if you look back at our original equation, it has π‘₯ minus π‘Ž and 𝑦 minus 𝑏. So therefore, it’s going be negative the values that we have inside of our parenthesis. So we’re gonna have values of negative three and negative nine because in our parenthesis we don’t have the negative sign, we have positive signs, so one to look out for when you do this kind of problem.

Okay, great! So we found the centre of the circle; we’ll try to find the radius. So we can do that because we know that in our circle, π‘Ÿ squared is equal to 64. So therefore, π‘Ÿ is gonna be equal to the square root of 64. So π‘Ÿ would be equal to eight. We’re gonna disregard the negative value for the square root of 64 because we aren’t actually talking about radius; we are talking about a length here. So there we have our final answer, which is the centre of our circle has π‘₯-coordinate negative three and 𝑦- coordinate negative nine and the radius of our circle is eight.

And a quick recap of what we have done, so first thing you need to do, you need to get it into the form of the equation of a circle. And to do that, we completed the square, then we simplified, and then we actually used the values that we found within the parenthesis to find our centre of the circle. And then we used our value on the right-hand side of the equation to find our radius.

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