Question

I need some help with math

Answer 1

Answer:

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

and

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

Step-by-step 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 $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ 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!