 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# The Circle - Problem 2

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Start by writing the given equation of a circle in standard form: r^2 = (x-h)^2 + (y-k)^2. In the equation, since r is being squared, the radius is the square root of the constant value. In other words, if r^2= 36, the radius is 6. To find the coordinate of the center, extract the h and k values. Since the quantity is (x-h), that means that the value of just "h" will have the opposite sign. In other words, if the quantity is (x-3), then h = 3. If the quantity is (x+5), then h = -5.

Finding the center and the radius from a given equation. So whenever we're given a equation for a circle, to find the information that's in there all we have to do is get it to a form we recognize. What I have here is similar to an equation for a circle, but I have these coefficients on my x and y² terms.

Our equation for circles can have those, so the first thing we have to do is get rid of those coefficients. In this case we just want to divide everything by 2 and our coefficients will be 1. So dividing through we end up with x minus 4 quantity squared plus y minus 1 quantity squared is equal to 72 over 2 or 36.

So now we just need to think back to our equation for a circle and extract the information. We're equal to 36 and that 36 is our radius squared which tells me that my, let's use a different color, radius is going to be equal to 6.

Our center is just extracted from the x minus and y minus terms and we're just dealing with x minus the x coordinate, so we're just dealing with x minus 4, that tells me my x coordinate is 4 and y minus the y coordinate tells me that I'm looking at 1 for my y coordinate of the center.

Another way you can look at is just think about what value needs to make this 0 and that will always be your coordinate for the center of the circle. So x is 4 to make this 0, y is 1 to make that 0, therefore 4,1 is our center.

So when finding the center and the radius of a equation, make sure it's in a form you recognize, make sure your coefficients are 1 and then just relate it to an equation for a circle x minus h² plus y minus k² is to r².