SOLUTION: Find the shortest distance from the origin to a point on the circle defined by x^2 + y^2 − 6x − 12y + 41 = 0
I got as far as (x-6)^2+(y-12)^2 = 5.
Algebra ->
Coordinate-system
-> SOLUTION: Find the shortest distance from the origin to a point on the circle defined by x^2 + y^2 − 6x − 12y + 41 = 0
I got as far as (x-6)^2+(y-12)^2 = 5.
Log On
Question 1114745: Find the shortest distance from the origin to a point on the circle defined by x^2 + y^2 − 6x − 12y + 41 = 0
I got as far as (x-6)^2+(y-12)^2 = 5. Found 2 solutions by greenestamps, ikleyn:Answer by greenestamps(13200) (Show Source):
You are not completing the square correctly. The standard form of the equation for the circle should have (x-3)^2 and (y-6)^2....
The circle has center (3,6) and radius 2.
The distance from the origin to the center of the circle, by the Pythagorean Theorem, is sqrt(45) = 3*sqrt(5); since the radius of the circle is 2, the shortest distance from the origin to a point on the circle is 3*sqrt(5)-2.