Help! Please classify the ff conics into circle, ellipse, parabola, hyperbola, or degenerate.
You should use the standard forms of equations of circle, ellipse, hyperbola or parabola, such that:
`(x - h)^2 + (y - k)^2 = r^2` (standard equation of circle, whose center is at `(h,k)` and its radius is r)
`y = a(x - h)^2 + k` (vertex form of equation of parabola, whose vertex is at `(h,k)` )
`x^2/a^2 + y^2/b^2 = 1` (standard equation of ellipse)
`x^2/a^2 - y^2/b^2 = 1` (standard equation of hyperbola)
Hence, having the standard forms as model, you need to try to convert each equation into its most suitable form.
Considering the equation `x^2+y^2-10x-4y-21=0` and comparing its form to each of the standard forms above, you may notice that the most suitable in this case seems to be the equation of the circle. The reasoning is made based on the number of terms that you can get expanding the squares from standard form of equation of circle.
Hence, all you need to do is to group the terms conveniently and then to complete the squares, such that:
`(x^2 - 10x) + (y^2 - 4y) - 21=0 `
`(x^2 - 10x + 25) + (y^2 - 4y + 4) - 25 - 4 - 21 = 0`
You should notice that you may complete the squares but you also need to preserve the equation, hence, once added the constants 25 and 4, you also need to subtract them.
`(x - 5)^2 + (y - 2)^2 = 50`
`(x - 5)^2 + (y - 2)^2 = (sqrt50)^2`
Hence, the equation `x^2+y^2-10x-4y-21=0` represents the general form of the equation of the circle whose center is at `(5,2)` and whose radius is of sqrt 50.
Considering the equation `x^2+y^2-4=0` , you may be tempted to compare it to the equation of ellipse since it contains `x^2` and `y^2` and no other members and the squares are added and not subtracted,as in case of hyperbola, but, since `a = b = 2` , the ellipse becomes a circle.
`x^2 + y^2 = 4 => x^2/4 + y^2/4 = 1`
Hence, `x^2 + y^2 = 4` represents the equation of a circle whose radius is `2` and whose center is at origin.