How do you find the center and radius for #21y^2+19x^2+83=4x−2x^2−84y#?

1 Answer
Aug 22, 2016

Answer:

The center of circle is #(2/21,-2)# and radius is #5/21#

Explanation:

The equation of a circle is of the type #x^2+y^2+2gx+2fy+c=0#, i.e. coefficients of #x# and #y# are equal and that of #xy# is #0#. In such a case coordinates of center are #(-g,-f)# and radius is given by #sqrt(g^2+f^2-c)#.

Here equation is given to be #21y^2+19x^2+83=4x-2x^2-84y# or

#21x^2+21y^2-4x+84y+83=0# or

#x^2+y^2-4/21x+4y+83/21=0#.

Hence center is #(2/21,-2)# and radius is

#sqrt((2/21)^2+2^2-83/21)#

= #sqrt(4/441+(84-83)/21)#

= #sqrt(4/441+1/21)#

= #sqrt(4/441+21/441)#

= #sqrt(25/441)=5/21#

Hence the center of circle is #(2/21,-2)# and radius is #5/21#