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What is the standard form of an equation of a circle?
Equation: (x-1)^2+(y+4)^2=16
Graph:
Let's start by recalling the standard form of an equation of a circle.
(x- h)^2+(y- k)^2= r^2
In this form, ( h, k) is the center of the circle and r is its radius. We are told that the center of the circle is ( 1, - 4) and its radius is 4. This information is enough to write the equation.
LHS-(x-1)^2=RHS-(x-1)^2
sqrt(LHS)=sqrt(RHS)
LHS-4=RHS-4
Now, we will make a table of values to find points on the curve. Be aware that the radicand cannot be negative!
| x | ± sqrt(16-(x-1)^2)-4 | y=sqrt(16-(x-1)^2)-4 | y=- sqrt(16-(x-1)^2)-4 |
|---|---|---|---|
| - 3 | ± sqrt(16-( - 3-1)^2)-4 | - 4 | - 4 |
| - 2 | ± sqrt(16-( - 2-1)^2)-4 | ≈ - 1.4 | ≈ - 6.6 |
| - 1 | ± sqrt(16-( - 1-1)^2)-4 | ≈ - 0.5 | ≈ - 7.5 |
| 0 | ± sqrt(16-( 0-1)^2)-4 | ≈ - 0.1 | ≈ - 7.9 |
| 1 | ± sqrt(16-( 1-1)^2)-4 | 0 | - 8 |
| 2 | ± sqrt(16-( 2-1)^2)-4 | ≈ - 0.1 | ≈ - 7.9 |
| 3 | ± sqrt(16-( 3-1)^2)-4 | ≈ - 0.5 | ≈ - 7.5 |
| 4 | ± sqrt(16-( 4-1)^2)-4 | ≈ - 1.4 | ≈ - 6.6 |
| 5 | ± sqrt(16-( 5-1)^2)-4 | - 4 | - 4 |
Finally, let's plot the points and connect them with a smooth curve.