How do you find the center, radius, x- and y- intercepts and graph of the circle : x^2+y^2-2x-4y-4=0?

1 Answer
Jul 9, 2015

Answer:

You convert the equation to standard form and use the values of #h# and #k# to calculate these values.

Explanation:

Step 1. Convert the equation to standard form.

The standard form for the equation is #(x-h)^2 + (y-k)^2 = r^2#.

We make the conversion by "completing the square".

#x^2 +y^2 -2x -4y -4 = 0#

#x^2 +y^2 -2x -4y = 4#

#(x^2-2x) + (y^2 -4y) = 4#

#(x^2-2x +1) -1 + (y^2 -4y +4) -4 = 4#

#(x-1)^2 -1 + (y-2)^2 -4 = 4#

#(x-1)^2 + (y-2)^2 = 4 +1 +4#

#(x-1)^2 + (y-2)^2 = 9#

Step 2. Locate the centre.

The centre is at (#h,k#) = (#x,y#) = (#1,2#).

Step 3. Calculate the radius.

The radius is #r = sqrt9 = 3#.

Step 4. Calculate the #y#-intercepts.

#(x-1)^2 + (y-2)^2 = 9#

Set #x = 0#.

#(0-1)^2 + (y-2)^2 = 9#

#1+ (y-2)^2 = 9#

#(y-2)^2 = 9 -1 = 8#

#y-2 = ±sqrt8 = ±2sqrt2#

#y = 2±2sqrt2#

The #y#-intercepts are at (#0,2-sqrt2#) and (#0,2+sqrt2#).

Step 4. Calculate the #x#-intercepts.

#(x-1)^2 + (y-2)^2 = 9#

Set #y = 0#.

#(x-1)^2 + (0-2)^2 = 9#

#(x-1)^2 + 4 = 9#

#(x-1)^2 = 9 -4#

#(x-1)^2 = 5#

#x-1 = ±sqrt5#

#x = 1 ±sqrt5#

The x-intercepts are at (#1-sqrt5,0#) and (#1+sqrt5,0#).