Answer:
[tex](y-7)^2+(x-2)^2=16[/tex]
and
[tex](x+2)^2+(y-15)^2 = 9[/tex]
Step-by-step explanation:
The standard equation of a circle is [tex](x-h)^2+(y-k)^2=r^2[/tex] where the coordinate (h,k) is the center of the circle.
Second Problem:
- We can start with the second problem which uses this info very easily.
- (h,k) in this problem is (-2,15) simply plug these into the equation. [tex](x--2)^2+(y-15)^2=r^2[/tex] .
- We can also add the radius 3 and square it so it becomes 9. The equation.
- This simplifies to [tex](x+2)^2+(y-15)^2 = 9[/tex].
First Problem:
- 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.
- [tex]y^2-14y+x^2-4x+37=0[/tex] first subtract 37 from both sides so the equation is now [tex]y^2-14y+x^2-4x=-37[/tex].
- [tex]y^2-14y+x^2-4x+37=0[/tex] by adding [tex](-\frac{b}{2a} )^2[/tex] to both the x and y portions of this equation you can complete the squares. [tex](-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2[/tex] and [tex](-\frac{-4}{2(1)})^2[/tex] which equals 49 and 4.
- Add 49 and 4 to both sides and the equation is now:[tex]y^2-14y+49+x^2-4x+4=-37+49+4[/tex] You can simplify the y and x portions of the equations into the perfect squares or factored form [tex](y-7)^2[/tex] and [tex](x-2)^2[/tex].
- Finally put the whole thing together. [tex](y-7)^2+(x-2)^2=16[/tex].
I hope this helps!
