# 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?

Nov 20, 2016

Set your compass for a radius of 20 units, place the center at $\left(10 , - 15\right)$, 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:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

In the given equation, $A = C \mathmr{and} B = 0$, therefore, the given equation describes a circle. The standard equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where, $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ 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:

$\left({x}^{2} - 20 x + {h}^{2}\right) + \left({y}^{2} + 30 y + {k}^{2}\right) - 75 = {h}^{2} + {k}^{2}$

Add 75 to both sides of the equation:

$\left({x}^{2} - 20 x + {h}^{2}\right) + \left({y}^{2} + 30 y + {k}^{2}\right) = {h}^{2} + {k}^{2} + 75$

Set the middle term in the right side of the pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, equal to the corresponding term in the equation:

$- 2 h x = - 20 x$

Solve for h:

$h = 10$

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

${\left(x - 10 x\right)}^{2} + \left({y}^{2} + 30 y + {k}^{2}\right) = {10}^{2} + {k}^{2} + 75$

Set the middle term in the right side of the pattern, ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$, equal to the corresponding term in the equation:

$- 2 k y = 30 y$

Solve for h:

$k = - 15$

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

${\left(x - 10 x\right)}^{2} + {\left(y - - 15\right)}^{2} = {10}^{2} + {\left(- 15\right)}^{2} + 75$

Combine the constants on the right:

${\left(x - 10 x\right)}^{2} + {\left(y - - 15\right)}^{2} = 400$

Express the constant as a square:

${\left(x - 10 x\right)}^{2} + {\left(y - - 15\right)}^{2} = {20}^{2}$

This is the equation of a circle with a center $\left(10 , - 15\right)$ and a radius of 20.

Nov 20, 2016

It is the equation of circle.

#### Explanation:

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

It is the equation of circle.

and it can be written as ${\left(x - 10\right)}^{2} + {\left(y + 15\right)}^{2} = 75 + 100 + 225 = 400 = {20}^{2}$

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

Also check here

As ${B}^{2} - 4 A C = 0 - 4 = - 4 < 0$, it is a circle.