How do you find the center and radius of the circle # x^2-2x+y^2-8y+1=0#?

2 Answers
Dec 19, 2016

Radius: #4#
Centre: #(-1, 4)#

Explanation:

You should convert to the form #(x - a)^2 + (y- b)^2 = r^2#. Complete the square twice to achieve this.

#1(x^2 - 2x + n) + 1(y^2 - 8y + m) = -1#

Our goal here is to make the expressions in parentheses into perfect squares. We can use the formula #n = (b/2)^2# and #m = (b/2)^2#, where #b# is the x-coefficent term.

So,

#n = (-2/2)^2 = 1#

#m = (-8/2)^2 = 16#

We need to add and subtract this number inside the parentheses to keep the equation equivalent.

#1(x^2 - 2x + 1 - 1) + 1(y^2 - 8x + 16 - 16) = -1#

#1(x^2 - 2x + 1) - 1 + 1(y^2 - 8x + 16) - 16 = -1#

#(x - 1)^2 + (y - 4)^2 = -1 + 17#

#(x + 1)^2 + (y - 4)^2 = 16#

The radius of a circle in the form #(x - a)^2 + (y - b)^2 = r^2# is #r#, so in this case #4#. The centre is at #(a, b)#, in this case #(-1, 4)#.

Hopefully this helps!

Dec 19, 2016

You complete the squares by adding #h^2 and k^2# to both sides of the equation. Please see the explanation.

Explanation:

The standard Cartesian form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2" [1]"#

where #(x, y)# is any point on the circle, #(h, k)# is the center, and #r# is the radius.

Equation [2] is the same as equation [1] but with the squares expanded and equation [3] is the given equation with some spaces added for missing terms:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#
#x^2 - 2x " " + y^2 - 8y" "+ 1= 0" [3]"#

Let's try to make equations [2] and [3] match by adding #h^2 and k^2# to both sides of equation [3] and label it equation [4]:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#
#x^2 - 2x + h^2 + y^2 - 8y + k^2 + 1= h^2 + k^2" [4]"#

Subtract 1 from both sides of equation [4] and label it equation [5]:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#
#x^2 - 2x + h^2 + y^2 - 8y + k^2 = h^2 + k^2 - 1" [5]"#

Now that equation [2] and equation [5] both have 6 terms on the left side, we can see that we can find the value of h by equating the second term in equation [2] with the same term in equation [5]:

#-2hx = -2x#

#h = 1#

Substitute 1 for h into equation [5] and number it equation [6]:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#
#x^2 - 2x + 1^2 + y^2 - 8y + k^2 = 1^2 + k^2 - 1" [6]"#

We can find the value of k by equating the fifth term in equation [2] with the same term in equation [5]:

#-2ky = -8y#

#k = 4#

Substitute 4 for k into equation [6] and number it equation [7]:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#
#x^2 - 2x + 1^2 + y^2 - 8y + 4^2 = 1^2 + 4^2 - 1" [7]"#

Because we used the patterns for #(x - h)^2 and (y - k)^2#, respectively, we know that the first three terms become #(x - 1)^2# and the next three terms become #(y - 4)^2#:

#(x - 1)^2 + (y - 4)^2 = 1^2 + 4^2 - 1" [8]"#

Simplify the right side:

#(x - 1)^2 + (y - 4)^2 = 4^2" [9]"#
#(x - h)^2 + (y - k)^2 = r^2" [1]"#

Comparing equation [9] with equation [1], we can see that the center is #(1, 4)# and the radius is 4.