Question

Please help me...

Problem:

The standard form of a circle is (x-h)2+(y-k)2=r2 and vertex form for the parabola, y-k=a(x-h)2. The (h,k) pair is the center of the circle and the vertex of the parabola. The radius of the circle is ‘r’ and the focal length of the parabola is f=1/(4a). For the following General Conic Equation: x2+y2-2x-4y-4=0 complete the following problems showing all your work:


A. Complete the square showing all your work to convert to Standard Form:


B. If this is a circle, state the coordinates of the center and give the radius. If this is a parabola, then state the coordinates of the vertex and give the focal length. Show all your work.


C. Sketch (or plot, or graph) the Conic. Label the point and values you found in part B. Also label the top-most, bottom-most, right-most, and left-most points on the circle or the focus and directrix on the parabola.

Answer 1

Part A)
`(x-1)^(2)+(y-2)^(2)=3^(2)`

Part B) The coordinates of the center are
`(h,k)=(1,2)` and the radius is equal to
`r=3\ units`

Part C) The graph in the attached figure

Explanation:

we have

`x^(2) +y^(2)-2x-4y-4=0`

Part A) Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`(x^(2)-2x) +(y^(2)-4y)=4`

Complete the square twice. Remember to balance the equation by adding the same constants to each side

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

`(x^(2)-2x+1) +(y^(2)-4y+4)=9`

Rewrite as perfect squares

`(x-1)^(2)+(y-2)^(2)=3^(2)` ------> Is a circle

Part B) If this is a circle, state the coordinates of the center and give the radius.

we know that

The equation of the circle in standard form is equal to

`(x-h)^(2)+(y-k)^(2)=r^(2)`

where

(h,k) is the center

r is the radius

In this problem we have

`(x-1)^(2)+(y-2)^(2)=3^(2)`

therefore

The coordinates of the center are
`(h,k)=(1,2)`

The radius is equal to
`r=3\ units`

Part C) Plot the circle

using a graphing tool

see the attached figure