Circles - Pre-Calculus

The standard form for the equation of a circle with radius , and centered at point is

.

Here, , so the equation is

.

Note: one way to think of this equation is to remember the Pythagorean Theorem.

If the center is at the origin then the equation is

.

This describes a right triangle for any x and y that satisfy this equation. Here r is the hypotenues, but when all values of x and y are used it stays the same and the points map out a circle with radius r.

The rules of graph translation apply in the same way as with any function. That is they move the origin in the opposite direction by a and/or b.

The center of the circle is .

Find the horizontal distance from the center to the edge of the circle. At the center , at the edge . The difference is . This is the radius.

Plug these values: into the standard form for the equation of a circle.

This gives

.

Remember that the standard form for the equation of a circle is given by the following formula:

Where the point (h,k) gives the center of the circle, and r is the radius. We can see from the form in which the equation is expressed in the problem that the only thing different with our form is that the terms on the left side of the equation are divided by 4. With some algebra, we'll multiply both sides by 4 to eliminate the 4's from the left side of equation:

Now we can see that our equation is the same as the formula for a circle in standard form, where (h,k) is (3,-2) and r=4.

First we must express the equation in standard form we can determine what the radius of our circle will be. The standard form for the equation of a circle is given as follows:

Where the point (h,k) gives the center of the circle and r is the radius of the circle, which can be easily determined by taking the square root of once the equation is in standard form. Our first step is to multiply both sides of the equation by 3 to cancel the division by 3 on the left side:

Now we can see that our equation is in standard form, where h=-5 and k=4, which tells us the coordinates of the center of the circle:

We can also determine the radius of the circle by taking the square root of :

Now that we know the center of the circle is at (-5,4), and that its radius is 3, we can find the points directly above and below the center by adding 3 to its y-coordinate, and then subtracting 3, respectively, giving us:

and

Similarly, to find the points directly to the left and to the right of the center, we subtract 3 from its x-coordinate, and then add 3, respectively, giving us:

and

Remember that in order for the equation of a circle to be in standard form, it must be written in the following way:

Where the point (h,k) tells us where the center of the circle is, and r is the radius of the circle. From our answer choices, we can see that the following is the only equation in which there are no fractions and there is addition of terms and not subtraction, which means it is in the standard form shown above:

For this circle, the center would be at (2,-3), and the radius would be 3.

We must begin by recalling the general formula for the equation of a circle.

Where circle has center of coordinates and radius of .

That means that looking at our equation, we can see that the center is .

If , then taking the square roots gives us a radius of 2.

We then look at our possible choices. Only two are centered at . Of these two, one has a radius of 2 while one has a radius of 4. We want the former.

Since the circle is centered at we use the most basic form for the equation for a circle:

.

Given the circle has a radius of marks, which represent units each, the circle has a radius of units.

We then plug in for :

and simplify: .

The equation for a circle in general is for a circle with center and radius . We know that the center is and that one of the points is , so we can determine by plugging these values in:

Now we can generalize the equation as

Recall that the equation of a circle is for the center and the radius.

In this case, we have as the center.

Note the negatives in the formula and be careful simpilfying.

When we are done, we have:

which gives us our answer when simplified.

To solve, simply use the formula for a circle as given below.

Thus, our answer is:

To find the standard form of this equation, we have to complete the square for both x and y. It's easiest to do this if we group together the y terms first, then the left terms, and subtract the constant from both sides:

original: subtract 1 from both sides; group x and y

To make the y terms into a square, we have to add 1, since half of 2 is 1, and .

To make the x terms into a square, we have to add 9, since half of -6 is -3, and :

Now we just have to re-write the y and x terms as the squares that they are, and simplify the right side:

To determine the center and radius, put this equation into standard form. Standard form is , where is the center and is the radius.

First, group the y terms together and the x terms together, and subtract 37 from both sides:

We're leaving a space after these terms because we're completing the square. To make the y terms into a square, we need to add 49, since half of -14 is -7, and

To make the x terms into a square, we need to add 4, since half of -4 is -2, and

We'll need to add these to both sides to keep both sides equal:

Now we can re-write the quadratics on the left as squares, and simplify the right side:

Interpreting this equation, we can see that the center would be at and the radius is

To write this in standard form, we will have to complete the square for both x and y. To do this more easily, group the x terms, then the y terms, and then subtract 43 from both sides:

To complete the square for x, we have to add to both sides; to complete the square for y, we have to add to both sides:

Condense the left side and add together the numbers on the right:

The point is the center of the circle - it is not on the circle.

We can test to see if the other points are actually on the circle by plugging in their x and y values into the equation. For example, to verify that

is actually on the circle, we can plug in for x and for y:

this is true, so that point is on the circle.

To quickly figure this out, we can plug in 5 for x and -2 for y and see what happens:

Since the value is greater than 9, this point is outside the radius of this circle.

The equation of a circle centered at with radius units in standard form is

For the circle ceentered at with radius units has the equation

The equation of a circle centered at with radius units in standard form is

For the circle ceentered at with radius units has the equation

or

The standard form of the equation of a circle is

where the center of the circle resides at the point .

Given the center of a circle and a point on the rim of the circle, one can use the distance formula to find the radius.

Now plug in the point for the center and radius into the standard equation of a circle:

To determine the standard-form equation, we'll have to complete the square for both x and y. It will be really helpful to re-group our terms to do that:

Adding 9 will complete the square for x, since

Adding 16 will complete the square for y, since

Now we just need to simplify. Re-write the left side as two binomials squared, and add the numbers on the right side: