Question

I need some help with math

#1 rewrite this in standard form, then state the center of the circle as an ordered pair and identify the radius. Show your work
y^(2)-14y+x^(2)-4x+37=0

#2 Write the equation of the circle with center (-2,15) and radius 3.

Answer 1

`(y-7)^2+(x-2)^2=16`

and

`(x+2)^2+(y-15)^2 = 9`

Explanation:

The standard equation of a circle is
`(x-h)^2+(y-k)^2=r^2` where the coordinate (h,k) is the center of the circle.

Second Problem:

1. We can start with the second problem which uses this info very easily.
2. (h,k) in this problem is (-2,15) simply plug these into the equation.
   `(x--2)^2+(y-15)^2=r^2` .
3. We can also add the radius 3 and square it so it becomes 9. The equation.
4. This simplifies to
   `(x+2)^2+(y-15)^2 = 9`.

First Problem:

1. The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
2. 
   `y^2-14y+x^2-4x+37=0` first subtract 37 from both sides so the equation is now
   `y^2-14y+x^2-4x=-37`.
3. 
   `y^2-14y+x^2-4x+37=0` by adding
   `(-(b)/(2a) )^2` to both the x and y portions of this equation you can complete the squares.
   `(-(b)/(2a))^2=(-(-14)/(2(1)))^2` and
   `(-(-4)/(2(1)))^2` which equals 49 and 4.
4. Add 49 and 4 to both sides and the equation is now:
   `y^2-14y+49+x^2-4x+4=-37+49+4` You can simplify the y and x portions of the equations into the perfect squares or factored form
   `(y-7)^2` and
   `(x-2)^2`.
5. Finally put the whole thing together.
   `(y-7)^2+(x-2)^2=16`.

I hope this helps!