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.

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 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....

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(52797)   (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 ! !

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SOLUTION: Find the shortest distance from the origin to a point on the circle defined by x^2 + y^2 &#8722; 6x &#8722; 12y + 41 = 0 I got as far as (x-6)^2+(y-12)^2 = 5.

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.