Question

What is the radius of a circle with the equation x^2+y^2+6x-2y+3?

Answer 1

Answer:

r = √13

Step-by-step explanation:

Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms:  x^2 + 6x            + y^2 -2y             = 3.  Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."  

We need to "complete the square" of x^2 + 6x.  I'll just jump in and do it:  Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x:     x^2 + 6x  + 3^2 - 3^2.  Then do the same for y^2 - 2y:  y^2 - 2y + 1^2 - 1^2.

Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2.  Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1.  Making these replacements:

(x + 3)^2 - 9 + (y - 1)^2 -1 = 3.  Move the constants -9 and -1 to the other side of the equation:  (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13

Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r:  r^2 = 13, so that the radius is r = √13.

Answer 2

Answer:

2.646

Step-by-step explanation:

Complete the square, put the equation in standard form, then evaluate the radius.

... x² + y² + 6x - 2y +3 = 0 . . . . given equation in general form

... x² +6x + 9 +y² -2y +1 = -3 +9 +1 . . . . add the squares of half the coefficients of the x- and y-terms; subtract 3

... (x +3)² +(y -1)² = 7 . . . . standard form equation for a circle

The term on the right, 7, is the square of the radius. so the radius is ...

... r = √7 ≈ 2.646