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?

2 Answers
Nov 20, 2016

Answer:

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.

Nov 20, 2016

Answer:

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.