All Precalculus Resources
Example Questions
Example Question #84 : Conic Sections
Write the equation for a circle with a center at (5,6) and a radius of 2.
The general equation for a circle is:
The coordinates of the center of the circle is (h,k) and r is the radius.
So plugging in the coordinates the circle equations is:
Example Question #1313 : Pre Calculus
Find the equation of the circle in standard form.
The equation of a circle in standard form is given as
where is the coordinates of the center of the circle and is the length of the radius.
In order to get the equation
in standard form we must complete the square.
First we group together similar variables
Then we complete the square by
- finding half of the coefficients of each variable with degree 1
- squaring those results
- and then adding those numbers to the equation.
As such,
And because the square of a difference is given as this equation through factoring
we have
or
Example Question #25 : Determine The Equation Of A Circle In Standard Form
Given the following equation for a circle, determine the coordinates of its center, as well as the coordinates of the four points directly, above, below, to the left, and to the right of the center:
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
Example Question #26 : Determine The Equation Of A Circle In Standard Form
Graph the circle indicated by the equation
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.
Example Question #31 : Determine The Equation Of A Circle In Standard Form
If each mark on the graph represents units, what is the equation of the circle?
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: .
Example Question #1316 : Pre Calculus
Which point is NOT on the circle defined by ?
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.
Example Question #31 : Circles
Which best describes the point and the circle ?
The point is the focus of the circle
The point is on the circle
The point has no relationship with the circle
The point is outside the circle
The point is inside the circle
The point is outside 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.
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