Question

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

Answer 1

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]}