12.2 Circles  (Page 2/2)

So, to understand the formula below, think of it as the $y=mx+b$ of circles.

Mathematical formula for a circle

From this, it is very easy to graph a circle in standard form.

Graphing a circle in standard form

Graph $(x+5{)}^2+(y–6{)}^2=10$	The problem. We recognize it as being a circle in standard form.
$h=\mathrm{–5}$ $k=6$ $r^2=10$	You can read these variables straight out of the equation, just as in $y=mx+b$ . Question: how can we make our equation’s $(x+5)$ look like the standard formula’s $(x-h)$ ? Answer: if $h=-5$ . In general, $h$ comes out the opposite sign from the number in the equation. Similarly, $(y-6)$ tells us that $k$ will be positive 6.
Center: (–5,6) Radius: $\sqrt{10}$	Now that we have the variables, we know everything we need to know about the circle.
	And we can graph it! $\sqrt{10}$ is, of course, just a little over 3—so we know where the circle begins and ends.

Got questions? Get instant answers now!

Just as you can go from a formula to a graph, you can also go the other way.

If a circle is given in nonstandard form , you can always recognize it by the following sign: it has both an $x^2$ and a $y^2$ term, and they have the same coefficient.

• $\mathrm{–3}{x}^2–3y^2+x–y=5$ is a circle: the $x^2$ and $y^2$ terms both have the coefficient –3
• $3x^2–3y^2+x–y=5$ is not a circle: the $x^2$ term has coefficient 3, and the $y^2$ has –3
• $3x^2+3y=5$ is not a circle: there is no $y^2$ term

Once you recognize it as a circle, you have to put it into the standard form for graphing. You do this by completing the square... twice!

Graphing a circle in nonstandard form

Graph $2x^2+2y^2–12x+28y–12=0$	The problem. The equation has both an $x^2$ and a $y^2$ term, and they have the same coefficient (a 2 in this case): this tells us it will graph as a circle.
$x^2+y^2–6x+14y–6=0$	Divide by the coefficient (the 2). Completing the square is always easiest without a coefficient in front of the squared tem.
$(x^2–6x)+(y^2+14y)=6$	Collect the $x$ terms together and the $y$ terms together, with the number on the other side.
$(x^2–6x+9)+(y^2+14y+49)=6+9+49$	Complete the square for both $x$ and $y$ .
$(x–3{)}^2+(y+7{)}^2=64$	Rewrite our perfect squares. We are now in the correct form. We can see that this is a circle with center at (3,–7) and radius 8. (*Remember How the signs change on $h$ and $k$ !)
	Once you have the center and radius, you can immediately draw the circle, as we did in the previous example.

Got questions? Get instant answers now!

Going from the definition of a circle to the formula

If you’re following all this, you’re now at the point where you understand the definition of a circle...and you understand the formula for a circle. But the two may seem entirely unconnected. In other words, when I said $(x–h{)}^2+(y–k{)}^2=r^2$ is the formula for a circle, you just had to take my word for it.

In fact, it is possible to start with the definition of a circle, and work from there to the formula, thus showing why the formula works the way it does.

Let’s go through this exercise with a specific example. Suppose we want to find the formula for the circle with center at (–2,1) and radius 3. We will start with the definition: this circle is the set of all the points that are exactly 3 units away from the point (–2,1). Think of it as a club. If a point is exactly 3 units away from (–2,1), it gets to join the club; if it is not exactly 3 units away, it doesn’t get to join.

You already know what the formula is going to be, but remember, in this exercise we’re not going to assume that formula—we’re going to assume nothing but the definition, and work our way to the formula. So here is our starting point, the definition for this circle:

“The distance from ( $x$ , $y$ ) to (–2,1) is 3.”

Any point ( $x$ , $y$ ) that meets this criterion is in our club. Using the distance formula that we developed above, we can immediately translate this English language definition into a mathematical formula. Recall that if xxxd is the distance between the points ( $x_1$ , $y_1$ ) and ( $x_2$ , $y_1$ ), then $(x_2-x_1{)}^2+(y_2-y_1{)}^2=d^2$ (Pythagorean Theorem). So in this particular case,

$(x+2{)}^2+(y-1{)}^2=9$

Note that this corresponds perfectly to the formula given above. In fact, if you repeat this exercise more generically—using ( $h$ , $k$ ) as the center instead of (–2,1), and r as the radius instead of 3—then you end up with the exact formula given above, $(x–h{)}^2+(y–k{)}^2=r^2$ .

For each of the remaining shapes, I’m going to repeat the pattern used here for the circle. First I will give the geometric definition and then the mathematical formula. However, I will not take the third step, of showing how the definition (with the distance formula) leads to the formula: you will do this, for each shape, in the exercises in the text.