Does anyone know how to solve #x^2+y^2+11x-13y=15# ?. Thank you so much for those who will answer

1 Answer
Jun 19, 2018

In standard form:

#(x-(-11/2))^2+(y-13/2)^2=(sqrt(350)/2)^2#

which is a circle with centre #(h, k) = (-11/2, 13/2)# and radius #r = sqrt(350)/2#

Explanation:

Given:

#x^2+y^2+11x-13y=15#

By completing the square for #x# and for #y# we can get the equation into standard form:

#(x-h)^2+(y-k)^2 = r^2#

where #(h, k)# is the centre of the circle and #r# the radius.

#15 = x^2+y^2+11x-13y#

#color(white)(15) = x^2+11x+(11/2)^2+y^2-13y+(13/2)^2-(11/2)^2-(13/2)^2#

#color(white)(15) = (x+11/2)^2+(y-13/2)^2-121/4-169/4#

#color(white)(15) = (x+11/2)^2+(y-13/2)^2-290/4#

Adding #290/4# to both ends and transposing, this becomes:

#(x+11/2)^2+(y-13/2)^2=350/4#

That is:

#(x-(-11/2))^2+(y-13/2)^2=(sqrt(350)/2)^2#

So this is a circle with centre #(h, k) = (-11/2, 13/2)# and radius #r = sqrt(350)/2#

graph{(x^2+y^2+11x-13y-15)((x+11/2)^2+(y-13/2)^2-0.1) = 0 [-26, 14, -3.68, 16.32]}