Question #de290

2 Answers
Jan 11, 2017

#(x-4)^2 +(y+5)^2 =36= 6^2

Explanation:

Rewrite as #x^2 -8x +y^2 +10y +5=0#

#x^2 -8x +16 -16 +y^2 +10y +25-25+5#

#(x-4)^2 +(y+5)^2 -16-25+5=0#

#(x-4)^2 +(y+5)^2 =36= 6^2#

Jan 11, 2017

You complete the squares, using the patterns:
#(x - h)^2 = x^2 - 2hx + h^2" [1]"#
and
#(y - k)^2 = y^2 - 2ky + k^2" [2]"#

Explanation:

The standard form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2" [3]"#

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

Given: #x^2 + y^2 - 8x + 10y + 5 = 0" [4]"#

Move the x terms together, the y terms together, and the constant to the right:

#x^2 - 8x + y^2 + 10y = -5" [4]"#

We want to make the first 3 terms in equation [4] look like the right side of equation [1] so we insert an #h^2# as the third term but, to keep the equation balanced, we must add #h^2# on the right:

#x^2 - 8x + h^2 + y^2 + 10y = h^2 -5" [5]"#

The first 3 terms of equation [5] look like the right side of equation [1].

We can match the #-2hx# in equation [1] with the #-8x# in equation [5] and write the equation:

#-2hx = -8x#

Find the value of h by dividing both sides of the equation by -2x:

#h = 4#

This means that, in equation [5], we can replace the terms #x^2 - 8x + h^2# with #(x - 4)^2# and replace the #h^2# on the right with 16:

#(x - 4)^2 + y^2 + 10y = 16 -5" [6]"#

We want to make the y terms in equation [6] look like the right side of equation [2} so we add a #k^2# on the left but, to keep the equation balanced, we must add a #k^2# to the right side:

#(x - 4)^2 + y^2 + 10y + k^2 = k^2 + 16 -5" [7]"#

The y terms on the left of equation [7] look like the right side of equation [2].

Match the #-2ky# in equation [2] with the #+10y# in equation [7] and write the equation:

#-2ky = +10y#

Find the value of k by dividing both sides by -2y:

#k = -5#

This means that, in equation [7], we can replace #y^2 + 10y + k^2# with #(y - -5)^2# and the k^2 on the right with 25:

#(x - 4)^2 + (y - -5)^2 = 25 + 16 -5" [8]"#

Simplify the constants on the right:

#(x - 4)^2 + (y - -5)^2 = 36" [9]"#

Write the constant as a square:

#(x - 4)^2 + (y - -5)^2 = 6^2" [10]"#

This is a circle with a radius of 6 and a center at #(4, -5)#