SOLUTION: Give the equation for the right half of the circle. C(2,3). I plotted it out to x=2+squrt9-(y-3)^2. Not sure about the signs if I have them right or not. thanks for checking.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Give the equation for the right half of the circle. C(2,3). I plotted it out to x=2+squrt9-(y-3)^2. Not sure about the signs if I have them right or not. thanks for checking.       Log On


   

Question 43230This question is from textbook algebra and trigonometry with analytic geometry
: Give the equation for the right half of the circle. C(2,3). I plotted it out to x=2+squrt9-(y-3)^2. Not sure about the signs if I have them right or not. thanks for checking.
 This question is from textbook algebra and trigonometry with analytic geometry

Found 2 solutions by psbhowmick, AnlytcPhil:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
You haven't specified the radius yet.
But, I think (looking at the algebraic expression you wrote) the radius is 3.
Then the equation of the circle with centre at (2,3) and radius equal to 3 units is
%28x-2%29%5E2+%2B+%28y-3%29%5E2+=+3%5E2_________(1)


As the equation of semi-circle which is the right half of the circle is reqd. so x+%3E=+2.
Simplifying (1) we have
%28x-2%29%5E2+=+3%5E2+-+%28y-3%29%5E2
or %28x-2%29%5E2+=+9+-+%28y%5E2+-+6y+%2B+9%29
or %28x-2%29%5E2+=+9+-+y%5E2+%2B+6y+-+9
or %28x-2%29%5E2+=+6y+-+y%5E2
or x+=+2+%2B-+sqrt%28+6y+-+y%5E2+%29 [square-rooting both sides and then adding 2 to both sides]

As 'x' has to be greater than or equal to 2, so x+=+2+%2B+sqrt%28+6y+-+y%5E2+%29 is the reqd equation to right half of the circle.

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Give the equation for the right half of the circle. C(2,3).
I plotted it out to x=2+squrt9-(y-3)^2.  Not sure about 
the signs if I have them right or not.  thanks for checking.

I will assume the radius is 3.  You forgot to give that.

The equation of the circle is

(x - 2)² + (y - 3)² = 3²

(x - 2)² + (y - 3)² = 9

Since you want the right half, 
solve for x.

(x - 2)² = 9 - (y - 3)²

Since you want the right half,
take positive square roots of
both sides:
            ____________
   x - 2 = Ö9 - (y - 3)² 
                ____________
       x = 2 + Ö9 - (y - 3)² 

Yes it looks like you're right.



Edwin