Exercise 13 Page 825

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. (x- 1)^2+(y-( - 4))^2= 4^2 ⇕ (x-1)^2+(y+4)^2=16 Now, let's draw our circle. To do so, we will isolate the y-variable.

(x-1)^2+(y+4)^2=16

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Solve for y

SubEqn

LHS-(x-1)^2=RHS-(x-1)^2

(y+4)^2=16-(x-1)^2

RadicalEqn

sqrt(LHS)=sqrt(RHS)

y+4=± sqrt(16-(x-1)^2)

SubEqn

LHS-4=RHS-4

y=± sqrt(16-(x-1)^2)-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.