Circle has the equation #x^2+y^2-6x+10y-15=0#, how do you graph the circle using the center (h,k) radius r?

2 Answers
Jul 9, 2018

Answer:

See explanation...

Explanation:

Given:

#x^2+y^2-6x+10y-15=0#

Rearrange into standard form for the equation of a circle as follows...

Complete the square for both #x# and #y#:

#0 = x^2+y^2-6x+10y-15#

#color(white)(0) = x^2-6x+9+y^2+10y+25-49#

#color(white)(0) = (x-3)^2+(y+5)^2-7^2#

Adding #7^2# to both ends and transposing, this becomes:

#(x-3)^2+(y+5)^2=7^2#

or if you prefer:

#(x-3)^2+(y-(-5))^2=7^2#

which is in the form:

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

  • the equation of a circle with centre #(h, k) = (3, -5)# and radius #r=7#.

graph{(x^2+y^2-6x+10y-15)((x-3)^2+(y+5)^2-0.08)=0 [-13.67, 17.33, -13, 2.5]}

Jul 9, 2018

Answer:

The center is #=(3,-5)# and the radius is #=7#

Explanation:

The standard equation of a circle, center #C=(h,k)# and radius #=r# is

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

Convert the given equation to the standard form by completing the square

#x^2+y^2-6x+10y-15=0#

#x^2-6x+y^2+10y=15#

#x^2-6x+9+y^2+10y+25=15+9+25#

#(x-3)^2+(y+5)^2=49=7^2#

The center is #=(3,-5)# and the radius is #=7#

The graph is as follows :

graph{x^2+y^2-6x+10y-15=0 [-14.8, 17.23, -12.3, 3.72]}