What is the center, radius, general form, and standard form of #x^2 + y^2 - 2x +6y - 3 =0#?

1 Answer
Dec 21, 2016

Answer:

General form is #(x-1)^2+(y+3)^2=(sqrt13)^2#.

This is the equation of a circle, whose center is #(1,-3)# and radius is #sqrt13#.

Explanation:

As there is no term in the quadratic equation #x^2+y^2-2x+6y-3=0# and coefficients of #x^2# and #y^2# are equal,

the equation represents a circle.

Let us complete the squares and see the results

#x^2+y^2-2x+6y-3=0#

#hArrx^2-2x+1^2+y^2+6y+3^2=1^2+3^2+3=13#

or #(x-1)^2+(y+3)^2=(sqrt13)^2#

It is the equation of a point which moves so that its distance from point #(1,-3)# is always #sqrt13# and hence equation represents a circle, whose radius is #sqrt13#.