How do you graph #x^2 + y^2 - 8x - 10y + 5=0#?

1 Answer
Shwetank Mauria
Feb 21, 2016

To graph the equation #x^2+y^2−8x−10y+5=0#, draw a circle with center at #(4,5)# and radius #6#.

Explanation:

As the equation #x^2+y^2−8x−10y+5=0# is of degree #2# and coefficients of #x^2# and #y^2# are equal and that of #xy# is #0#, it is clearly an equation of a circle.

Let us convert it into sum of squares as follows

#(x^2−8x+16)+(y^2−10y+25)=16+25-5# or

#(x-4)^2+(y-5)^2=36# or

#(x-4)^2+(y-5)^2=6^2#

which is the equation of a circle with center at #(4,5)# and radius #6#.

Hence, to graph the equation #x^2+y^2−8x−10y+5=0#, draw a circle with center at #(4,5)# and radius #6#.

graph{(x^2+y^2-8x-10y+5)(x^2+y^2-8x-10y+40.95)=0 [-10, 18, -2, 12]}

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