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Course: Algebra (all content)>Unit 17

Lesson 3: Standard equation of a circle

Writing standard equation of a circle

Given a circle on the coordinate plane, Sal finds its standard equation, which is an equation in the form (x-a)²+(y-b)²=r².

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• Around , he explains that we need to use the pythagorean theory to find the radius r. But can't we just estimate the no. of units from the centre (point -1, 1) to the point (7.5, 1), which also lies on the circumference?
• sometimes you will need precision in your answers. Pythagorean Theorem will give you as much precision as you need
• How do you put the square root of (#-#)squared + (#-#)squared, into a calculator? Or how do you solve this without a calculator?
• Some calculators are really simple and can't find exponents or square roots, but others can. You make a square root symbol, and then fit all of (#-#)^2 + (#-#)^2 inside of it. If you don't have a calculator that can do that, I guess you can just do it out with pencil and paper. Subtract #-# inside the parentheses, then use long multiplication to "square" the result (just multiply by itself), and then.... I think there's a way to find square roots on paper, but you might have to find an online calculator.
• At , Sal makes a right triangle, do you have to make a right triangle, can't you just measure straight up, down, to the right or to the left to find the radius because every point on a circle is the same distance from the center?
• You can do it that way. Sal is simply showing another way of doing it.
• At , He said that -5 squared is 25. Isn't it -25?
(1 vote)
• No, (-5)^2 is 25 because -5 * -5 is 25. Hope this helps!
• The equation gets 74 at the end, can someone tell me what 74 is?
• Since all the other parts of the equation are squared, we have to find the radius squared, giving us 74. The radius itself would just be the square root of 74.
• what if the given were the endpoints of a diameter? how can you solve it?
• Use the distance formula to find the length of the diameter, and then divide by 2 to get the radius. Then find the midpoint of the diameter which will be the center of the circle. Now you have the coordinates of the center and the radius and that is all that is necessary to write the standard equation of the circle.
• can you explain the relationship of this equation:
x=x*y^2
i'm getting all kinds of weirds stuffs trying to solve
• If x≠0, you can divide both sides by x to get 1=y², which means y=±1. If x=0, this equation is true no matter what y is.

The second part tells us that the entire y-axis will be part of this graph. The first part also includes two horizontal lines at y=±1. So the graph will look something like the ± sign if you drew the vertical line all the way through.
• I get so confused in this really helps but I still don't understand can someone please help?
(1 vote)
• First you need to know that the equation for a circle is (x-a)^2 + (y-b)^2 = r^2 where the center is at point (a,b) and the radius is r. so for instance (x-2)^2 + (y-3)^2 = 4 would have the center at (2,3) and have a radius of 2 since 4 = 2^2. This means if you went a distance of 2 away from that center point you would be on the circle. Some super simple ones are adding 2 to the x or y coordinate. so the center at (2,3) means (2+2, 3), (2-2, 3), (2, 3+2) and (2, 3-2) are on the circle.

Now the video starts giving a circle, where we know one point on the circle. Thankfully it tells us the center, so we can handle the first half of the equation. (x-a)^2 + (y-b)^2 where a = -1 ad b = 1 makes it (x+1)^2 + (b-1)^2. So easy. The radius isn't so easy.

there are no points directly above, below or to the left or right of the center you can gauge the radius from. In fact you only know for sure of one point since it tells you. Well, the radius is the distance from the center of the circle to the edge, and the other point we know is by defiition on the edge. Now, how do you find the distance between two points.

If you imagine the two points had a horizontal and vertical line connecting them, this would make a right triangle. Can you manage from there?
• Hey! I have a doubt, why does in this case what we first called (h,k) as in (x-h)^2 + (y-k)^2 = r^2, (-1,1) (the center) while in the video "features of a circle from its standard equation" was what we now used as (6,-4).
(1 vote)
Center is at (-1,1) = the point (h,k)
A point on circle is (6,-4) = one of many points (x,y)
Sal uses these values to determine the radius of the circle by using the standard form of a circle and plugging in the given values
[6-(-1)]^2 + (-4-1)^2 = r^2
Do the calculations.
7^2 + (-5)^2 = r^2
49 + 25 = r^2
74 = r^2
Take the square root to find "r".
r = sqrt(74)

Hope this helps.
• WHY is 'a' and 'b' subtracted in the standard equation?