SOLUTION: A circle has an equation x^2+y^2-3x-4y+4=0 Graph the circle using the center (h,k) and radius r. Find the intercepts, if any, of the graph.

Algebra ->  Rational-functions -> SOLUTION: A circle has an equation x^2+y^2-3x-4y+4=0 Graph the circle using the center (h,k) and radius r. Find the intercepts, if any, of the graph.      Log On


   

Question 970939: A circle has an equation
x^2+y^2-3x-4y+4=0
Graph the circle using the center (h,k) and radius r. Find the intercepts, if any, of the graph.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the Squares, and make use of the standard form of the finished equation.

x%5E2-3x%2By%5E2-y%2B4=0-----THE MISTAKE HERE-.
x%5E2-3x%2B%283%2F2%29%5E2%2By%5E2-y%2B%281%2F2%29%5E2%2B4-%283%2F2%29%5E2-%281%2F2%29%5E2=0
%28x-3%2F2%29%5E2%2B%28y-1%2F2%29%5E2%2B4-9%2F4-1%2F4=0
%28x-3%2F2%29%5E2%2B%28y-1%2F2%29%5E2=10%2F4-4
%28x-3%2F2%29%5E2%2B%28y-1%2F2%29%5E2=-3%2F2------This will not work. This is not a real circle.
r%5E2=-3%2F2---No good.

Look for any mistakes in your given equation or in my steps. I found none in mine but examine if you want.
-
EDIT: FOUND THE MISTAKE IN MY WORK.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
A circle has an equation
x^2+y^2-3x-4y+4=0
Graph the circle using the center (h,k) and radius r. Find the intercepts, if any, of the graph.
x%5E2+%2B+y%5E2+-+3x+-+4y+%2B+4+=+0

x%5E2+-+3x+%2B+y%5E2+-+4y+=+-+4

%28x%5E2+-+3x%29+%2B+%28y%5E2+-+4y%29+=+-+4 ------- Grouping variables

 

%28x%5E2+-+3x+%2B+9%2F4%29+%2B+%28y%5E2+-+4y+%2B+4%29+=+-+4+%2B+9%2F4+%2B+4

%28x+-+3%2F2%29%5E2+%2B+%28y+-+2%29%5E2+=+9%2F4

Since the center-radius form of the equation of a circle is: %28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2,

it's obvious that the center, or (h, k) = (3%2F2, 2)

Since r%5E2+=+9%2F4, then r, or radius = sqrt%289%2F4%29, or 3%2F2

Based on this info, you should be able to graph the circle.