Equation for a Tangent Circle on the Y-Axis at (3,7)

  • Thread starter Thread starter john560
  • Start date Start date
  • Tags Tags
    Circle
AI Thread Summary
To find the equation of a circle centered on the y-axis and tangent to a vertical line at (3,7), the correct form is derived from the general circle equation. The initial attempt mistakenly presented a parabola instead of a circle. The accurate equation should be (x^2)/9 + (y-7)^2/9 = 1, which can also be expressed as (x^2)/3^2 + (y-7)^2/3^2 = 1 for clarity. The discussion emphasizes the importance of correctly applying the circle formula and ensuring the equation is in standard form. Proper notation and structure are crucial for accurately representing the circle's properties.
john560
Messages
6
Reaction score
0

Homework Statement


Write an equation for the circle centered on the y-axis and is tangent to a vertical line at the point (3,7)


Homework Equations





The Attempt at a Solution



(x^2)/9 + (y-7)/9
 
Physics news on Phys.org
john560 said:

Homework Statement


Write an equation for the circle centered on the y-axis and is tangent to a vertical line at the point (3,7)


Homework Equations





The Attempt at a Solution



(x^2)/9 + (y-7)/9

Sorry to sound rude, but what the hell kind of attempt is that? There is so much wrong with it that... ahh what's the point...

First start with the basics. Do you know the general formula for a circle with centre (h,k) and radius r?
 
Yes, I do know the basic formula for a circle.
 
Right my apologies, you just missed the (y-7)2 and didn't make it equal to 1.
 
welcome to pf!

hi john560! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
john560 said:
Write an equation for the circle centered on the y-axis and is tangent to a vertical line at the point (3,7)

(x^2)/9 + (y-7)/9

i] that isn't an equation, is it? :redface:

ii] anyway, it looks like a parabola, not a circle :wink:

try again :smile:
 
Yea i forgot to square the Y side and make it equal to one
Mentallic said:
Right my apologies, you just missed the (y-7)2 and didn't make it equal to 1.

(x2)/9 + (y-7)2/9 =1
 
john560 said:
Yea i forgot to square the Y side

(x2)/9 + (y-72)/9 =1
Not quite. This is what I'm seeing:
\frac{x^2}{9}+\frac{y-7^2}{9}=1
Is that what you really mean? (Look carefully at the 2nd fraction.)
 
john560 said:
Yea i forgot to square the Y side and make it equal to one





(x2)/9 + (y-72)/9 =1

Yup! :biggrin:

(except the 2 is the wrong side of the bracket, and some professors would prefer you to multiply throughout by 9 :wink:)
 
<br /> \frac{x^2}{9} + \frac{(y-7)^2}{9}=1<br />
 
  • #10
Yep, that's it :smile:

But like tiny-tim said, it is usually more appropriate to have the circle equation in the form (x-h)^2+(y-k)^2=r^2 rather than \frac{(x-h)^2}{r^2}+\frac{(y-k)^2}{r^2}=1
 
  • #11
tiny-tim said:
(except the 2 is the wrong side of the bracket, and some professors would prefer you to multiply throughout by 9 :wink:)

So that's what he meant<br /> \frac{x^2}{3^2} + \frac{(y-7)^2}{3^2}=1<br />

Thank you guys, wish this site had a thumbs up icon so i could show my appreciation that way.
 
Last edited:
  • #12
now how did i do that? :rolleyes:
 
Back
Top