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Mathematics 20 Online
OpenStudy (anonymous):

what is the radius of this equation: x squared plus y squared minus 14x plus 6y plus 42=0

OpenStudy (anonymous):

4

OpenStudy (anonymous):

Ok after completing the square you get the equations (x-7)^2+(x+3)^2=16. Take the square root of 16 which is 4. That is your radius.

OpenStudy (anonymous):

Ok after completing the square you get the equations (x-7)^2+(x+3)^2=16. Take the square root of 16 which is 4. That is your radius.

OpenStudy (anonymous):

x^2 + y^2 - 14x + 6y +4 = 0 x^2-14x-y^2+6y = -4 (x^2-14x + 49)-49 - (y^2-6y+9)-9 = -4 (x^2 - 14x + 49) - (y^2 - 6y + 9) = -4 + 49 + 9

OpenStudy (anonymous):

agree with helpp1234

OpenStudy (anonymous):

centre of circle is (-g,-f) in circle equation x^2 +y^2 +2gx+2fy+c=0 and radius =sqrt(g^2 +f^2 -c)

OpenStudy (anonymous):

so radius = 4

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