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) About Me  (Show Source):
You can put this solution on YOUR website!


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....

x%5E2+%2B+y%5E2+-+6x+-+12y+%2B+41+=+0
x%5E2-6x%2B9+%2B+y%5E2-12y%2B36+=+-41%2B9%2B36
%28x-3%29%5E2+%2B+%28y-6%29%5E2+=+4

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.

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.

                You really need to learn these two techniques:


1)  how to complete the square,   and

2)  how to apply it to get the standard equation of a circle from its general equation.


For it,  read the lessons in this site
    - HOW TO complete the square - Learning by examples

    - Standard equation of a circle
    - General equation of a circle
    - Transform general equation of a circle to the standard form by completing the squares
    - Identify elements of a circle given by its general equation


                    H a p p y   l e a r n i n g  ! !