Hello me and my friend have been trying to figure out whay is wrong with my ciricle formula. When completing a check to see if the formula is correct we get the wrong answer even though we should be edning up with the correct answer. It goes like this:

(the math done here is for my algbra two and trig class)

Circle:

Parent fucntion: x^{2 }+ y^{2 }= r^{2 }

a= 9 b= 9

h=1 k=4

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

Transformation: (x,y) = (9x+1), (9y+4)

Equation: y= 4 plus or minus 9 square root 9-(x-1)^{2 }

Check:

Point: (x,y) = (1,13)

(1-1)^{2} + (y-4)^{2} = 9

0^{2 }+ (y-4)^{2 }= 9

0 + square root (y-4)^{2 }= square root 9

y-4 = 3

that would equal 7 which does not equal 13. However if I were not to square root 9 I would get 13, But that would be breaking the rules, Can anyone explain what I am doing wrong here?

Guest Nov 26, 2017

#1**+1 **

I'm no expert at this, but notice that you have:

x^2 + y^2 = r^2 as the "parent function", then further down you have this:

(x-1)^2 + (y-4)^2 = 9 Notice that you are squaring the LHS. Shouldn't you then square the RHS as well? So, when checking it you have this:

(1-1)^2 + (y-4)^2 = 9, but if you squared 9^2 on RHS, then you will have:

y =-5 and **y = 13, which is what you want.**

Guest Nov 26, 2017

#2**0 **

I am not squaring the right side of my equatin as It is already squared. I don't need to put three squared unlike the left side where I need to do some steps before squaring. The right side is already solved. If you rerfer to the parent function x^{2 }+ y^{2 }= r2 you'll see that r is squared. Sorry for the confusion.

Guest Nov 26, 2017

edited by
Guest
Nov 26, 2017

#3**+1 **

I believe I figured it out if you use (x-1)^{2} + (y-4)^{2} = 9 and plug in 1 for x and 13 for y you will get 3 = 3.

As the x part of the equation equals zero then 13-4 equals 9 sqeared. So you end up with 9^{2 }= 9. Then just square root them and you get 3 equals 3.

Guest Nov 26, 2017

#4**+2 **

I don't know what your " a " and " b " stand for, but it looks like you want the equation of a circle with its center at (1, 4) that passes through (1, 13) .

the radius of the circle = the distance between the center and a point on the circle

the radius of the circle = the distance between (1, 4) and (1, 13)

Notice how the (1, 4) and (1, 13) both have the same x coordinate.

So the distance between the two points is just 13 - 4 .

the radius of the circle = 13 - 4 = 9

So we know that r = 9 , h = 1 , and k = 4 .

Plugging these values directly into the standard form of a circle.....

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

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

(x - 1)^{2} + (y - 4)^{2} = 81 This is the equation of the circle, and here's a graph of it.

hectictar Nov 26, 2017

#5**+1 **

Thank you what I was looking for was to find the error in my checking of the equation you just wrote. Though I have now figured that out but thank you for your resonse. A stands for the amount dilated horzintoally and b stand for the amount dilated vertically.

Can someone please close this forum/check that it has been answer?

Guest Nov 26, 2017