Video: Graphing Circles

Which of the following circles is the one of the equation (π‘₯ βˆ’ 3)Β² + (𝑦 + 1)Β² = 4?

03:44

Video Transcript

Which of the following circles is the one of the equation π‘₯ minus three squared plus 𝑦 plus one squared is equal to four?

To answer this question, we need to remember the center-radius form of the equation of a circle. If a circle has its center at the point with coordinates β„Ž, π‘˜ and a radius of π‘Ÿ units, then its equation in center-radius form is given by π‘₯ minus β„Ž squared plus 𝑦 minus π‘˜ squared is equal to π‘Ÿ squared.

We can see that the structure of this is identical to the structure of the equation that we’ve been given. So, we need to compare the two equations to determine the values of β„Ž, π‘˜, and π‘Ÿ. We’ll then be able to work out the center and radius of our circle. And so, we’ll be able to determine what it looks like when plotted.

Comparing the first brackets then, we have π‘₯ minus β„Ž in the general form and π‘₯ minus three in our specific equation. So, we see that the value of β„Ž, the π‘₯-coordinate of the center of the circle, is three.

Now, comparing the second brackets, we have 𝑦 minus π‘˜ in the general form and 𝑦 plus one in our equation. So, we see that negative π‘˜ is equal to one. To find the value of π‘˜, we need to multiply or divide both sides of this equation by negative one, which gives that π‘˜ is equal to negative one. So, the 𝑦-coordinate of the center is negative one.

Finally, comparing the right-hand sides of our equations, we have that π‘Ÿ squared is equal to four. To find the value of π‘Ÿ, we need to take the square root of each side of this equation. And as four is a square number, we should recognise that its square root is just the integer two. Now normally when we solve an equation by square rooting, we would take plus or minus the square root. But as π‘Ÿ is the radius of a circle, it needs to take a positive value. So, we’re only going to take the positive square root of four.

By comparing the equation of our circle with the center-radius form then, we’ve found that our circle has a center at the point three, negative one and a radius of two units. We can now look at the graph we’ve been given to determine which is the correct plot for this circle. Here is the point three, negative one. And we can see that we have two circles centered on this point, the circles C and D. We need to determine which of these circles has the correct radius.

We can do this by drawing a line from this center to any point on the circumference of each circle. For simplicity, it makes sense to use a horizontal or vertical radius, like the one I’ve drawn here. The length of this line will be the difference in the π‘₯-coordinates of its two endpoints, which is three and one. And therefore, the length of this line, and the radius of the circle C, is two units. Just for comparison’s sake, we can see that the radius of the larger circle, the blue circle, is four units, because its radius is the difference between the π‘₯-values of seven and three.

So, the circle, which has a center at the point three, negative one and a radius of two units and is, therefore, the circle representing the equation π‘₯ minus three squared plus 𝑦 plus one squared equals four, is circle C.

Notice that circle D represents a common misconception. It has the correct center but an incorrect radius. In fact, the radius that’s been drawn is the value we would get if we had forgotten to square root. We know that the radius squared is four, but the radius itself is two. This is a common misconception or a common mistake to make. The correct answer, when we remember to take the square root of the value on the right-hand side of the circle given in center-radius form, is circle C.

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