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Sia

Question
Math
Posted 2 months ago

A certain circle can be represented by the following equation.
x2+y220x+8y+115=0
x^{2}+y^{2}-20 x+8 y+115=0


What is the center of this circle?
\square , \square
What is the radius of this circle?
\square units
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
To find the center of the circle, we need to complete the square for the xx and yy terms in the equation
step 2
The given equation is x2+y220x+8y+115=0x^2 + y^2 - 20x + 8y + 115 = 0. We group the xx terms and the yy terms: (x220x)+(y2+8y)+115=0(x^2 - 20x) + (y^2 + 8y) + 115 = 0
step 3
To complete the square for the xx terms, we take half of the coefficient of xx, which is 20-20, and square it, giving (202)2=100(\frac{-20}{2})^2 = 100. We add and subtract 100100 inside the equation
step 4
To complete the square for the yy terms, we take half of the coefficient of yy, which is 88, and square it, giving (82)2=16(\frac{8}{2})^2 = 16. We add and subtract 1616 inside the equation
step 5
The equation now looks like this: (x220x+100)+(y2+8y+16)+11510016=0(x^2 - 20x + 100) + (y^2 + 8y + 16) + 115 - 100 - 16 = 0
step 6
We can rewrite the equation as (x10)2+(y+4)21=0(x - 10)^2 + (y + 4)^2 - 1 = 0
step 7
Rearranging the equation, we get (x10)2+(y+4)2=1(x - 10)^2 + (y + 4)^2 = 1
step 8
The center of the circle is at the point (h,k)(h, k) where hh and kk are the values that complete the square, so the center is (10,4)(10, -4)
step 9
The radius of the circle is the square root of the constant on the right side of the equation, which is 1=1\sqrt{1} = 1
[question 1] Answer
The center of the circle is (10,4)(10, -4).
The radius of the circle is 11 unit.
Key Concept
Completing the square to find the center and radius of a circle
Explanation
The center of a circle in the equation (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2 is (h,k)(h, k), and the radius is rr. By completing the square, we can rewrite the given equation in this form to find the center and radius.

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