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## High school geometry

### Course: High school geometry>Unit 7

Lesson 3: Expanded equation of a circle

# Circle equation review

Review the standard and expanded forms of circle equations, and solve problems concerning them.

## What is the standard equation of a circle?

left parenthesis, x, minus, start color #11accd, h, end color #11accd, right parenthesis, squared, plus, left parenthesis, y, minus, start color #ca337c, k, end color #ca337c, right parenthesis, squared, equals, start color #e07d10, r, end color #e07d10, squared
This is the general standard equation for the circle centered at left parenthesis, start color #11accd, h, end color #11accd, comma, start color #ca337c, k, end color #ca337c, right parenthesis with radius start color #e07d10, r, end color #e07d10.
Circles can also be given in expanded form, which is simply the result of expanding the binomial squares in the standard form and combining like terms.
For example, the equation of the circle centered at left parenthesis, start color #11accd, 1, end color #11accd, comma, start color #ca337c, 2, end color #ca337c, right parenthesis with radius start color #e07d10, 3, end color #e07d10 is left parenthesis, x, minus, start color #11accd, 1, end color #11accd, right parenthesis, squared, plus, left parenthesis, y, minus, start color #ca337c, 2, end color #ca337c, right parenthesis, squared, equals, start color #e07d10, 3, end color #e07d10, squared. This is its expanded equation:
\begin{aligned} (x-\blueD 1)^2+(y-\maroonD 2)^2&=\goldD 3^2 \\\\ (x^2-2x+1)+(y^2-4y+4)&=9 \\\\ x^2+y^2-2x-4y-4&=0 \end{aligned}

## Practice set 1: Using the standard equation of circles

Problem 1.1
• Current
left parenthesis, x, plus, 4, right parenthesis, squared, plus, left parenthesis, y, minus, 6, right parenthesis, squared, equals, 48
What is the center of the circle?
left parenthesis
comma
right parenthesis
units

Want to try more problems like this? Check out this exercise and this exercise.

## Practice set 2: Writing circle equations

Problem 2.1
• Current
A circle has a radius of square root of, 13, end square root units and is centered at left parenthesis, minus, 9, point, 3, comma, 4, point, 1, right parenthesis.
Write the equation of this circle.

Want to try more problems like this? Check out this exercise.

## Practice set 3: Using the expanded equation of circles

To interpret the expanded equation of a circle, we should rewrite it in standard form using the method of "completing the square."
Consider, for example, the process of rewriting the expanded equation x, squared, plus, y, squared, plus, 18, x, plus, 14, y, plus, 105, equals, 0 in standard form:
\begin{aligned} x^2+y^2+18x+14y+105&=0 \\\\ x^2+y^2+18x+14y&=-105 \\\\ (x^2+18x)+(y^2+14y)&=-105 \\\\ (x^2+18x\redD{+81})+(y^2+14y\blueD{+49})&=-105\redD{+81}\blueD{+49} \\\\ (x+\redD9)^2+(y+\blueD7)^2&=25 \\\\ (x-(-9))^2+(y-(-7))^2&=5^2 \end{aligned}
Now we can tell that the center of the circle is left parenthesis, minus, 9, comma, minus, 7, right parenthesis and the radius is 5.
Problem 3.1
• Current
x, squared, plus, y, squared, minus, 10, x, minus, 16, y, plus, 53, equals, 0
What is the center of this circle?
left parenthesis
comma
right parenthesis
What is the radius of this circle?
units

Want to try more problems like this? Check out this exercise and this exercise.

## Want to join the conversation?

• What should be done to adjust the radius of the circle in the practice? I could move the centre, but I'm lost as to how the radius should be adjusted.
• Move your cursor outwards slowly until the center of the circle is minimized. Then you can click and drag to adjust the radius of the circle.
It is a bit tricky when the circle is small but it can be done
• Parabolas are called Quadratics. What are circle equations called? Are they also called Quadratics because of the square? Thanks.
• Parabolas, hyperbolas, ellipses, and circles (which are just special ellipses) are collectively called "Conic Sections", since you can find each of these curves as the intersection between an infinite cone and a plane.

"Quadratic" refers to a polynomial of degree 2, and hence only describes parabolas.

Equations that describe circles don't really have a special name.
• In problem 2.2 how did you get √17?
• I think the graph is deceiving in that it is hard to read accurately.
At first glance it looks like the circle goes though the point (9,0) and therefore has a radius of 4.
However, upon closer looking, the circumference of the circle is just a bit beyond (9,0) and actually hits the point (9,-1).
That's why (using the Pythagorean theorem) the distance for the radius is calculated between the center (5,0) and the point (9,-1) which lies on its circumference. (Note that the point (9,1) could have been used instead as could (1,1) or (1,-1).)
• How can you find square roots without a calculator?
• If you don't need an exact answer, you can use linear approximations. Suppose we are trying to find the square root of a number, x, and we know the square root of a number, k, that is "close" to x. (k will likely be perfect square as it makes the math easier). The square root of x is approximately equal to (x + k)/(2 * sqrt(k)).
• Find the centre and radius of a circle whose equation is x2+y2-4y-21
• What you have written isn't an equation because it has no equals sign, so presumably you mean x^2 + y^2 - 4y - 21 = 0
We know that the general equation for a circle is ( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius.
So add 21 to both sides to get the constant term to the righthand side of the equation.
x^2 + y^2 -4y = 21
Then complete the square for the y terms.
x^2 + y^2 - 4y + 4 = 21 + 4
Then factor.
x^2 + ( y - 2 )^2 = 5^2
So, the center is at ( 0, 2 ) and the radius is 5.
• Hi! I really need an urgent response towards how should I find the equation of circle if the only given were the center (-1,-7) and a point (2,11).

Well, I tried doing using the standard equation of the circle and imputed the given sample and tried to square it. Unfortunately, WileyPLUS (Its a book converted into sort of online stuffs and other than pen and paper activities its all made online) as part of our school marked my answers wrong... can anyone help me? I really need to perfect it.
• The standard equation for a circle centred at (h,k) with radius r
is (x-h)^2 + (y-k)^2 = r^2
So your equation starts as ( x + 1 )^2 + ( y + 7 )^2 = r^2
Next, substitute the values of the given point (2 for x and 11 for y), getting
3^2 + 18^2 = r^2,
so r^2 = 333.
The final equation is (x+1)^2 + (y+7)^2 = 333
Hope this helps!
• What if the equation is 9x^2 + 24xy + 16y^2 + 90x - 130y = 0 ? I can't figure this out. It says that I should rotate the axes to remove the xy, but I can't really get it.
• A=9, B=24, C=16 so the discriminant is 576 - 4(9)(16) = 576 - 576 = 0. Therefore this is a parabola. Factorising the first three terms we get (3x+4y)^2 + 90x - 130y = 0. Let u = 3x+4y and v = -4x+3y. Then x = (3u - 4v)/25 and y = (4u + 3v)/25. So the equation becomes:

u^2 + (18/5) (3u-4v) + (26/5) (4u+3v) = u^2 + 158u/5 + 6v/5 = 0

Complete the square: (u+79/5)^2 - 6241/25 + 6v/5 = 0

Rearrange for v: v = 6241/30 - 5/6 (u+79/5)^2

So the directrix is v = 6241/30 + 6/20 = 625/3
And the focus is (u,v) = (-79/5, 6241/30 - 6/20) = (-79/5, 3116/15)

Writing in terms of x and y:
The directrix is -12x+9y = 625
And the focus is (x,y) = (-527/15, 112/5)