FIND THE EQATION OF CIRCLE CONCENTRIC WITH THE CIRCLE X2+Y2-2X-6Y+4=0 AND HAVING RADIUS 7?
First we write the equation of the circle
x² + y² - 2x - 6y + 4 = 0
In the form
(x - h)² + (y-k)² = r²
x² + y² - 2x - 6y + 4 = 0
Rearrange so that the x terms and the y terms are together
and the constant term on the right.
x² - 2x + y² - 6y = -4
Place the x terms and the y terms in parentheses
(x² - 2x) + (y² - 6y) = -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)(1/2) = -1
(-6)(1/2) = -3
2. Square the results:
(-1)² = +1
(-3)² = +9
3. Add at the end of each parentheses and to the opposite side.
(x² - 2x + 1) + (y² - 6y + 9) = -4 + 1 + 9
Factor each parenthetical expression as the square of a binomial on the left
and combine the terms on the right:
(x² - 2x + 1) = (x - 1)(x - 1) = (x - 1)²
(y² - 6y + 9) = (y - 3)(y - 3) = (x - 3)²
(x - 1)² + (y - 3)² = 6
Compare this to
(x - h)² + (y - k)² = r²
h=1, k = 3, r² = 6, so r = √6
The center = (h,k) = (1, 3) and the radius is √6
The desired circle has the same center (1,3) but a different radius,
7.
So the left side of the equation of the circle you wish to find
will be the same as the left side of the circle you were given,
but the right side will be r² = 7² or r = 49 instead of 6.
Answer: (x - 1)² + (y - 3)² = 49
The green circle is the circle you were given, and the red circle
is the one you were asked to find.
Edwin
