Let's use the distance formula to solve for the value of k:
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
We know that the point (a,k) is 5 units away from the point (5,6), so we can set up the equation:
5 = √[(a - 5)^2 + (k - 6)^2]
Squaring both sides, we get:
25 = (a - 5)^2 + (k - 6)^2
Expanding and simplifying:
25 = a^2 - 10a + 25 + k^2 - 12k + 36
Combining like terms:
0 = a^2 - 10a + k^2 - 12k + 36
Rearranging and factoring:
0 = (a^2 - 10a) + (k^2 - 12k + 36)
0 = (a - 5)^2 + (k - 6)^2 - 1
(a - 5)^2 + (k - 6)^2 = 1
This is the equation of a circle centered at (5,6) with a radius of 1. The value of k can be found by solving for it when a point on the circle is given. Since there are two points on the circle that are 5 units away from (5,6), we have two possible values for k:
When a = 6:
(a - 5)^2 + (k - 6)^2 = 1
(6 - 5)^2 + (k - 6)^2 = 1
1 + (k - 6)^2 = 1
(k - 6)^2 = 0
k - 6 = 0
k = 6
When a = 4:
(a - 5)^2 + (k - 6)^2 = 1
(4 - 5)^2 + (k - 6)^2 = 1
1 + (k - 6)^2 = 1
(k - 6)^2 = 0
k - 6 = 0
k = 6
Therefore, the value of k in (a,k) that is 5 units away from (5,6) is k = 6.
Answers & Comments
Let's use the distance formula to solve for the value of k:
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
We know that the point (a,k) is 5 units away from the point (5,6), so we can set up the equation:
5 = √[(a - 5)^2 + (k - 6)^2]
Squaring both sides, we get:
25 = (a - 5)^2 + (k - 6)^2
Expanding and simplifying:
25 = a^2 - 10a + 25 + k^2 - 12k + 36
Combining like terms:
0 = a^2 - 10a + k^2 - 12k + 36
Rearranging and factoring:
0 = (a^2 - 10a) + (k^2 - 12k + 36)
0 = (a - 5)^2 + (k - 6)^2 - 1
(a - 5)^2 + (k - 6)^2 = 1
This is the equation of a circle centered at (5,6) with a radius of 1. The value of k can be found by solving for it when a point on the circle is given. Since there are two points on the circle that are 5 units away from (5,6), we have two possible values for k:
When a = 6:
(a - 5)^2 + (k - 6)^2 = 1
(6 - 5)^2 + (k - 6)^2 = 1
1 + (k - 6)^2 = 1
(k - 6)^2 = 0
k - 6 = 0
k = 6
When a = 4:
(a - 5)^2 + (k - 6)^2 = 1
(4 - 5)^2 + (k - 6)^2 = 1
1 + (k - 6)^2 = 1
(k - 6)^2 = 0
k - 6 = 0
k = 6
Therefore, the value of k in (a,k) that is 5 units away from (5,6) is k = 6.