Unit 4, lesson 36

Circles, Semicircles, and Choosing a Branch

Where the circle equation comes from, why a full circle is not the graph of a function, and how to split it into upper and lower semicircle functions with exact domains and ranges.

  • Theory part 71
  • Practice part 72
  • Comparing Curves

A circle is about as simple a shape as there is, yet it is not the graph of any function, because most vertical lines that meet it meet it in two points. Calculus wants functions. The way out is to cut the circle into a top half and a bottom half, each of which is a perfectly good function. Choosing the right half, with the right domain and range, is the skill here.

Key terms

Circle, center and radius

Definition Let (h,k)(h, k) be a point in the plane and r>0r > 0 a real number. The circle with center (h,k)(h, k) and radius rr is the set of all points (x,y)(x, y) whose distance from (h,k)(h, k) is exactly rr. The center itself is not on the circle, and the points inside are not on it either.

In plain words Fix a point and a length. The circle is every point that sits exactly that far from the fixed point.

Example The circle with center (2,−1)(2, -1) and radius 33 contains (5,−1)(5, -1), three units to the right of the center, and (2,2)(2, 2), three units above it. The center (2,−1)(2, -1) is not a point of the circle.

Pythagorean relationship

Definition The distance between points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is

(x2−x1)2+(y2−y1)2.\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

It comes from the Pythagorean theorem applied to the right triangle whose legs are the horizontal distance ∣x2−x1∣\lvert x_2 - x_1 \rvert and the vertical distance ∣y2−y1∣\lvert y_2 - y_1 \rvert, and whose hypotenuse is the segment between the points.

In plain words Go across, then up. The straight line distance is the hypotenuse of the triangle you just walked along.

Example From the origin to (4,3)(4, 3) the legs are 44 and 33, so the distance is

42+32=25=5.\sqrt{4^2 + 3^2} = \sqrt{25} = 5.

So (4,3)(4, 3) lies on the circle of radius 55 centered at the origin.

Circle equation

Definition A point (x,y)(x, y) lies on the circle with center (h,k)(h, k) and radius r>0r > 0 exactly when

(x−h)2+(y−k)2=r2.(x - h)^2 + (y - k)^2 = r^2.

This is the distance condition with both sides squared, which is reversible here because both the distance and rr are nonnegative. The equation describes the curve implicitly, as a relation between xx and yy rather than as a formula for yy.

In plain words Horizontal offset squared plus vertical offset squared equals radius squared.

Example The equation x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0 is a circle in disguise. Completing the square in xx and in yy gives

(x2−6x+9)+(y2+4y+4)=12+9+4,\begin{aligned} &(x^2 - 6x + 9) + (y^2 + 4y + 4) \\[0.4em] &\quad = 12 + 9 + 4, \end{aligned}

that is (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25. The center is (3,−2)(3, -2) and the radius is 55.

Principal square root

Definition For a real number a≥0a \ge 0, the principal square root a\sqrt{a} is the unique real number b≥0b \ge 0 with b2=ab^2 = a. For a<0a < 0 there is no real square root. The equation u2=au^2 = a with a>0a > 0 has the two solutions u=au = \sqrt{a} and u=−au = -\sqrt{a}, and in general u2=∣u∣\sqrt{u^2} = \lvert u \rvert.

In plain words The radical sign always means the nonnegative root. If you want the negative one, you have to write the minus sign yourself.

Example The equation u2=9u^2 = 9 has solutions 33 and −3-3, but 9=3\sqrt{9} = 3 only. Likewise (−4)2=16=4\sqrt{(-4)^2} = \sqrt{16} = 4, which is ∣−4∣\lvert -4 \rvert, not −4-4.

Branch

Definition Solving the circle equation for yy gives (y−k)2=r2−(x−h)2(y - k)^2 = r^2 - (x - h)^2, so

y=k+r2−(x−h)2ory=k−r2−(x−h)2.\begin{gathered} y = k + \sqrt{r^2 - (x - h)^2} \\[0.4em] \text{or} \\[0.4em] y = k - \sqrt{r^2 - (x - h)^2}. \end{gathered}

Each choice of sign gives a function, called a branch of the circle. The radical is a nonnegative distance from the horizontal line y=ky = k, and the sign in front of it chooses the side. The complete circle with r>0r > 0 is not the graph of a function of xx, since the vertical line x=hx = h meets it at the two points (h,k+r)(h, k + r) and (h,k−r)(h, k - r).

In plain words One formula with a plus or minus sign is really two functions. Pick one sign and you get one half of the circle.

Example On the circle x2+y2=25x^2 + y^2 = 25, the input x=3x = 3 gives y2=16y^2 = 16, so y=4y = 4 or y=−4y = -4. The point (3,4)(3, 4) is on the branch with the plus sign and (3,−4)(3, -4) is on the branch with the minus sign.

Upper semicircle

Definition The upper semicircle of the circle with center (h,k)(h, k) and radius r>0r > 0 is the graph of

y=k+r2−(x−h)2,h−r≤x≤h+r.\begin{gathered} y = k + \sqrt{r^2 - (x - h)^2}, \\[0.4em] h - r \le x \le h + r. \end{gathered}

It is the part of the circle with y≥ky \ge k, including the two endpoints (h−r,k)(h - r, k) and (h+r,k)(h + r, k).

In plain words Take the plus sign. You get the top half, from the leftmost point of the circle over the top to the rightmost point.

Example For the centered circle x2+y2=25x^2 + y^2 = 25 the upper semicircle is y=25−x2y = \sqrt{25 - x^2}.

xx−5-5−3-3003355
yy0044554400

Lower semicircle

Definition The lower semicircle of the same circle is the graph of

y=k−r2−(x−h)2,h−r≤x≤h+r.\begin{gathered} y = k - \sqrt{r^2 - (x - h)^2}, \\[0.4em] h - r \le x \le h + r. \end{gathered}

It is the part of the circle with y≤ky \le k, and it shares the two endpoints (h±r,k)(h \pm r, k) with the upper semicircle.

In plain words Take the minus sign. You get the bottom half, a mirror image of the top half across the horizontal line through the center.

Example For x2+y2=25x^2 + y^2 = 25 the lower semicircle is y=−25−x2y = -\sqrt{25 - x^2}. At x=−3x = -3 it gives −4-4, at x=0x = 0 it gives −5-5, the lowest point of the circle, and at x=5x = 5 it gives 00.

Domain of a branch

Definition Both branches have the closed domain [h−r,h+r][h - r, h + r]. The radical is real exactly when r2−(x−h)2≥0r^2 - (x - h)^2 \ge 0, that is when ∣x−h∣≤r\lvert x - h \rvert \le r. The endpoints are included, because at x=h±rx = h \pm r the radicand is 00 and the root is 00.

In plain words The circle reaches rr units left and right of the center and no farther, so those are the allowed inputs.

Example For the shifted circle (x−1)2+(y−2)2=9(x - 1)^2 + (y - 2)^2 = 9 the center is (1,2)(1, 2) and the radius is 33. The upper branch is

y=2+9−(x−1)2,y = 2 + \sqrt{9 - (x - 1)^2},

with domain [−2,4][-2, 4]. At x=5x = 5 the radicand would be 9−16=−79 - 16 = -7, so 55 is not an allowed input.

Range of a branch

Definition As xx runs over [h−r,h+r][h - r, h + r], the radical r2−(x−h)2\sqrt{r^2 - (x - h)^2} takes every value from 00, at the two endpoints, up to rr, at x=hx = h. Hence the upper branch has range [k,k+r][k, k + r] and the lower branch has range [k−r,k][k - r, k]. Both ranges are closed intervals.

In plain words The top half rises from the height of the center to the top of the circle, and the bottom half drops from the height of the center to the bottom.

Example For the upper branch y=2+9−(x−1)2y = 2 + \sqrt{9 - (x - 1)^2} above, the endpoints (−2,2)(-2, 2) and (4,2)(4, 2) give the smallest output 22, and the center input x=1x = 1 gives the largest output 2+3=52 + 3 = 5. The range is [2,5][2, 5]. The lower branch of the same circle has range [−1,2][-1, 2].

Check yourself

  1. Write the two branches of x2+y2=16x^2 + y^2 = 16 as functions with their domains, and evaluate the lower branch at x=2x = 2.
Show the answer

The branches are y=16−x2y = \sqrt{16 - x^2} and y=−16−x2y = -\sqrt{16 - x^2}, both with domain [−4,4][-4, 4]. The lower branch at x=2x = 2 gives −12=−23-\sqrt{12} = -2\sqrt{3}.

  1. A circle has center (−3,1)(-3, 1) and radius 22. Give its upper branch with domain and range, and find its value at x=−2x = -2.
Show the answer

The upper branch is

y=1+4−(x+3)2,y = 1 + \sqrt{4 - (x + 3)^2},

with domain [−5,−1][-5, -1] and range [1,3][1, 3]. At x=−2x = -2 it gives 1+31 + \sqrt{3}.

  1. Show that (1,−3)(1, -3) lies on the circle x2+(y+1)2=5x^2 + (y + 1)^2 = 5 and decide which branch it belongs to.
Show the answer

Substituting gives 1+4=51 + 4 = 5, so the point is on the circle. The center is (0,−1)(0, -1), and −3<−1-3 < -1, so the point is below the center and lies on the lower branch y=−1−5−x2y = -1 - \sqrt{5 - x^2}. Indeed that branch gives −1−2=−3-1 - 2 = -3 at x=1x = 1.

How this lesson connects