Circles

The answer is C. Among other formulas, the CLT provides you with the circumference formula of $2\pi r$ and the area formula of $\pi r^2$. We need to use both in this case. First, we set the circumference formula equal to $64\pi$: $\pi r^2=64\pi$. Cancel the $\pi$ on both sides and we get $r^2=64$, so $r=8$.

We now plug the radius of $8$ into the circumference formula: $2\times 8\times \pi =16\pi$. Note that the CLT will virtually always leave $\pi$ in the answer choices. They know you don’t have a calculator and don’t expect you to calculate $\pi$!

The answer is B. To understand how the circle’s area could be equal to its circumference, we set the two formulas equal to each other in an equation: $\pi r^2=2\pi r$. After canceling the $\pi$ on both sides, we have $r^2=2r$. Although we might be tempted to take the square root of both sides, that’s not algebraically helpful because it will put the term $2r$ under a square root.

Instead, we’ll follow our principle for quadratics (anything with a variable raised to the second as the highest power in the expression): set quadratic equations equal to zero. Doing so and factoring yields the following:

$$\begin{gathered} r^2-2r=0 \\ r(r-2)=0 \\ r=0 \\ OR \\ r=2 \end{gathered}$$

Since it’s nonsensical to have a radius of zero units, the answer must be $2$.

The answer is A. The principle of dimensions is so important that we include it in detail in the Strategy Insights for this lesson. Put briefly, because a radius is a one-dimensional concept but area is a two-dimensional concept, they are not affected in the same way. We can discover this by actually calculating the area of circle $B$: $\pi \times 5^2=25\pi$. Circle $A$, we are told, has an area four times that large, so the area of Circle $A$ must be $100\pi$. What, then, is the radius of Circle $A$? It must be the square of $100$, which is $10$.

Note the trap answer here! Choice $C$ is very tempting because $20$ is four times as much as $5$. But the principle of dimensions tells us that’s a trap answer; it will always be a trap answer in questions like this.

The answer is D. We find here another example of a circle interacting with another shape, so we follow the strategy of determining what line segment the two shapes have in common. A square inscribed in a circle looks like this:

The answer is D. As usual, you should draw the figure on your scratch paper to envision the situation. The UnCLES method reminds us to observe right away what the question is asking for; the fact that we are looking for the part not painted helps our strategy right away. If the painted portion is equal to $50°$, then the unpainted part must be equal to $360-50=310°$.

We can use the arc length formula once we have the radius. If the distance across the field is $360$, then the radius must be $180$. Plugging $180$ and $310$ in the formula looks like this:

$$\begin{array}{rl}\frac{310}{360}& (2\times \pi \times 180)\\ \frac{31}{36}& (360\pi )\\ (31)& (\frac{360\pi }{36})\\ (31)& (10\pi )\\ & 310\pi \end{array}$$

Note how we rearranged the factors to divide $360$ by $36$, since that results in the integer value $10$. Working these problems with a calculator and with limited time requires that you seek ways to streamline the process. Also, although you can certainly memorize the arc and sector formulas, it also helps to realize that they both involve using the arc measure to find the fraction of the circle you’re dealing with. The only difference from there is that arc length is that fraction times the circumference while sector area is that fraction times the area.

The answer is B. This is a complicated question that invokes 1) the skill of completing the square; 2) the circle equation $[(x-h{)}^2+(y-k{)}^2=r^2]$; and 3) the fact that a tangent line is perpendicular to the radius at the point where the two lines intersect.

We begin by seeking to interpret the circle based on the circle equation, but notably the equation is not yet in the proper form. This is where completing the square comes in. We’ll begin with the y terms and return to x later. We can copy down $y^2-6y$, but we’re used to seeing a third term in this expression in order to factor it. What is that third term? We know from the form of the circle equation that we are looking for a perfect square quadratic, something that will factor into the form $(y-k{)}^2$. Completing the square involves two steps:

1. Divide the coefficient of the first-degree term (the term raised to the first power) in half;
2. Square the result.

The coefficient in this case is $6$, so by dividing it in half and squaring it, we get $9$. We add the $9$ after our first two terms to make $y^2-6y+9$. But we can’t just change an equation; that would unbalance it! So we add $9$ to the other side as well, making $25$. $y^2-6y+9$ can be factored into $(y-3)(y-3)$ or, better, $(y-3{)}^2$. That gives us the desired form $(y-k{)}^2$. Since that process is challenging to remember and repeat, here it is again, line by line:

$$\begin{array}{rl}{x}^2+(y^2-6y)& =16\\ x^2+(y^2-6y+9)& =16+9\\ x^2+(y-3{)}^2& =25\end{array}$$

But what about $x$? Why are there no other terms besides $x^2$? Whenever you see a missing term in an equation, you can assume there must be a zero involved; indeed, we have here the equivalent of $(x-0{)}^2$. Therefore, we now have sufficient information to identify the center $(h,k)$ of the circle: it is $(0,3)$.

We now bring in the property concerning the tangent line and the radius being perpendicular. To find the slope of the radius, we can use the center of the circle [$(0,3)$] and the point we know is on the circle [$(3,7)$]. The slope formula tells us that the slope of this line is $(7-3)/(3-0)=\frac{4}{3}$. Since perpendicular lines have a negative reciprocal slope, the answer must be $-\frac{3}{4}$.