Question

How do you identity if the equation `x^2+y^2-20x+30y-75=0` is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Answer 1

Set your compass for a radius of 20 units, place the center at `(10,-15)`, and then draw a circle. Please see the explanation for details and a reference.

Explanation:

Here is a reference Conic section - General Cartesian form that tells how to identitfy any equation of the form:

`Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`

In the given equation, `A = C and B = 0`, therefore, the given equation describes a circle. The standard equation of a circle is:

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

where, `(x, y)` is any point on the circle, `(h,k)` is the center, and r is the radius.

Add `h^2 + k^2` to both sides of the equation and group the x and y terms together:

`(x^2 - 20x + h^2) + (y^2 + 30y + k^2) - 75 = h^2 + k^2`

Add 75 to both sides of the equation:

`(x^2 - 20x + h^2) + (y^2 + 30y + k^2) = h^2 + k^2 + 75`

Set the middle term in the right side of the pattern, `(x - h)^2 = x^2 -2hx + h^2`, equal to the corresponding term in the equation:

`-2hx = -20x`

Solve for h:

`h = 10`

This allows us to substitute `(x - 10)^2` on the left and 10 for h on the right:

`(x - 10x)^2 + (y^2 + 30y + k^2) = 10^2 + k^2 + 75`

Set the middle term in the right side of the pattern, `(y - k)^2 = y^2 -2ky + k^2`, equal to the corresponding term in the equation:

`-2ky = 30y`

Solve for h:

`k = -15`

This allows us to substitute `(y - -15)^2` on the left and -15 for k on the right:

`(x - 10x)^2 + (y - -15)^2 = 10^2 + (-15)^2 + 75`

Combine the constants on the right:

`(x - 10x)^2 + (y - -15)^2 = 400`

Express the constant as a square:

`(x - 10x)^2 + (y - -15)^2 = 20^2`

This is the equation of a circle with a center `(10, -15)` and a radius of 20.

Answer 2

It is the equation of circle.

Explanation:

As the coefficients of `x^2` and `y^2` are equal and their is no term containing `xy` (i.r. coefficient of `xy` is `0`)

It is the equation of circle.

and it can be written as `(x-10)^2+(y+15)^2=75+100+225=400=20^2`

Hence this is an equation of a circle with center as `(10,-15)` and radius `20` and o can draw a circle of radius `20` with `(10,-15)` as center.
graph{x^2+y^2-20x+30y-75=0 [-31.17, 48.83, -34.88, 5.12]}

Also check here

As `B^2-4AC=0-4=-4<0`, it is a circle.