# Question Video: Finding the Center and Radius of a Circle by Completing the Square Mathematics • 11th Grade

By completing the square, find the center and radius of the circle π₯Β² + 6π₯ + π¦Β² β 4π¦ + 8 = 0.

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### Video Transcript

By completing the square, find the center and radius of the circle π₯ squared plus six π₯ plus π¦ squared minus four π¦ plus eight is equal to zero.

Weβre given the equation of a circle, and we can see this is given in the general form. We need to find the center and radius of this circle. To do this, letβs recall where we get the general form of the equation of a circle from. The standard form for the equation of a circle centered at the point π, πΏ with radius π is π₯ minus π all squared plus π¦ minus πΏ all squared is equal to π squared. So if weβre given a circle in this form, we can find its center and its radius just from the equation.

In the general form for the equation of a circle, we distribute these squares over our parentheses. Distributing the exponent over our first set of parentheses, either by using the FOIL method or binomial expansion, we get π₯ squared minus two ππ₯ plus π squared. Similarly, distributing the exponent over our second set of parentheses, we get π¦ squared minus two πΏπ¦ plus πΏ squared. And of course, this entire equation is equal to π squared. But at this point, we could simplify. π is just a constant, so we could just call negative two π π. Similarly, πΏ is also a constant, so we could call negative two πΏ π. Finally, we can combine the constants π squared, πΏ squared, and π squared into one constant weβll call π.

And this then gives us the general form for the equation of a circle: π₯ squared plus ππ₯ plus π¦ squared plus ππ¦ plus π is equal to zero. But weβre given the general form for the equation of a circle, and we need to get back to the standard form for the equation of a circle. And to do this, we see instead of squaring π₯, we want to square π₯ minus π. And instead of squaring π¦, we want to square π¦ minus πΏ. And this is exactly what we mean by using completing the square. Letβs now see how we can use completing the square to help us go from the general equation of a circle to the standard equation for a circle.

Remember that by completing the square, we can write the quadratic π₯ squared plus ππ₯ plus π as π₯ plus π over two all squared minus π squared over four plus π. This is because if we distribute the square over our parentheses on the right-hand side of this equation, either by using the FOIL method or binomial expansion, we get π₯ squared plus two times π over two multiplied by π₯ plus π squared over four minus π squared over four plus π. And of course, we can simplify. Two divided by two is equal to one, and π squared over four minus π squared over four is equal to zero. And of course, this is equal to our original quadratic.

However, we need to notice how weβve rewritten our polynomial. We have π₯ plus some constant all squared, and then we add on some other constant. This is exactly the form we needed to rewrite our equation in the standard form. So letβs now start completing the square on the general equation of a circle given to us in the question. Weβll start with just the terms which contain π₯. Thatβs π₯ squared plus six π₯. To complete the square on this, we first need to divide our coefficient of π₯ by two, and we know six divided by two is equal to three. So weβll start with π₯ plus three all squared. But if we were to stop here, when we distribute the square over our parentheses, we would get π₯ squared plus six π₯ plus nine. So to balance both sides of this equation, we need to subtract a constant nine.

Weβll now do exactly the same with the terms which contain π¦. Thatβs π¦ squared minus four π¦. And weβll do exactly the same thing we did before. We need to divide our coefficient of π¦, which is negative four, by two. And, of course, negative four divided by two is equal to negative two. So weβre going to square π¦ minus two. This time, if we distribute the square over our parentheses, we get π¦ squared minus four π¦ plus four. So to balance both sides of the equation, weβre going to need to subtract four.

We can now use both of these expressions to rewrite the general form of the equation of a circle in the standard form. First, weβve shown that π₯ squared plus six π₯ is equal to π₯ plus three all squared minus nine. So weβll write this into our equation. We can do exactly the same with π¦ squared minus four π¦. We can replace this with π¦ minus two all squared minus four. This gives us π₯ plus three all squared minus nine plus π¦ minus two all squared minus four plus eight is equal to zero. And now, this is almost in the general form for the equation of a circle. We just need to simplify.

We have negative nine minus four plus eight is equal to negative five. And remember, the radius of our circle is written on the other side of our equation. So we need to add five to both sides of this equation. This gives us π₯ plus three all squared plus π¦ minus two all squared is equal to five. And of course, we know that our radius π is usually squared. So instead of writing five, we could write the square root of five all squared. And now, this is exactly written in the standard form for the equation of a circle. We can find our values of π, πΏ, and π.

We have π is equal to negative three, πΏ is equal to two, and π is equal to the square root of five. And we know from the standard form for the equation of a circle, this means that our circle has center negative three, two and a radius of the square root of five. Therefore, by completing the square twice, we were able to find the center and radius of the circle π₯ squared plus six π₯ plus π¦ squared minus four π¦ plus eight is equal to zero. We got that the center of this circle was negative three, two and the radius of this circle was the square root of five.