write the equation of the circle satisfying the given condition:
concentric with the circle 4x^2 + 4y^2 - 8x + 4y + 1=0 and tangent to the line
y= -3/2.
First we write the equation of the circle
4x² + 4y² - 8x + 4y + 1 = 0
In the form
(x - h)² + (y-k)² = r²
4x² + 4y² - 8x + 4y + 1 = 0
Divide thru by 4, the common coefficient of x² and y²
x² + y² - 2x + y + 1/4 = 0
Rearrange so that the x terms and the y terms are together
and the constant term on the right.
x² - 2x + y² + y = -1/4
Place the x terms and the y terms in parentheses
(x² - 2x) + (y² + y) = -1/4
Complete the square in both parentheses by the rule
1. Multiply the coefficient of the second term in each parentheses by 1/2
2. Square the result
3. Add at the end of each parentheses and to the opposite side.
(x² - 2x + 1) + (y² + y + 1/4) = -1/4 + 1 + 1/4
Factor each parenthetical expression as the square of a binomial on the left
and combine the terms on the right:
(x - 1)² + (y + 1/2)² = 1
Compare this to
(x - h)² + (y - k)² = r²
h=1, k = -1/2, r² = 1, so r = 1
The center = (h,k) = (1, -1/2) abd the radius is 1
The desired circle has the same center (1, -1/2) but a different radius.
It is to be tangent to the line y = -3/2 which is a horizontal line.
The point (1, -3/2) is the point on the line y = -3/2 which is
directly below the center (1, -1/2), thus this point, (1, -3/2) must
be the point of tangency.
The radius of the desired circle is the distance between the center (1, -1/2)
and the point of tangency (1, -3/2) directly below it. This distance is
(-1/2) - (-3/2) = -1/2 + 3/2 = 2/2 = 1.
But 1 is the radius of the given circle!!!!!! So the given circle
4x² + 4y² - 8x + 4y + 1 = 0 is ITSELF tangent to the line y = -3/2.
Thus there is something wrong with your problem unless your teacher allows
you to say that a circle is concentric with ITSELF!!!
Edwin
AnlytcPhil@aol.com
