How do you write the equation #x^2+y^2+8x-10y-1=0# in standard form and find the center and radius?

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Answer 1 · 2018-02-11T09:24:12

Standard form of equation of a circle is #(x+4)^2+(y-5)^2=(sqrt42)^2#. Center is #(-4,5)# and radius is #sqrt42#.

#x^2+y^2+8x-10y-1=0# can be written as

#x^2+8x+y^2-10y=1#

or #(x^2+8x+16)+(y^2-10y+25)=1+16+25#

or #(x+4)^2+(y-5)^2=42#

or #(x+4)^2+(y-5)^2=(sqrt42)^2#

This is the standard form of equation of a circle and

indicates that the point #(x,y)# moves so that its distance from the point #(-4,5)# is always #sqrt42#

Hence, center of circle is #(-4,5)# and radius is #sqrt42#

graph{x^2+y^2+8x-10y-1=0 [-18, 10, -2, 12]}