Help me please...
1 answer:
We have that
x²+y²-4x-6y-12=0
Group terms that contain the same variable, and move the constant to the opposite side of the equation
(x²-4x)+(y²-6y)=12
Complete the square twice. Remember to balance the equation by adding the same constants to each side
(x²-4x+4)+(y²-6y+9)=12+4+9
Rewrite as perfect squares
(x-2)²+(y-3)²=25
the answer part A) is
(x-2)²+(y-3)²=5²-----> this is the standard form of the equation of a circle
Part B) (x-2)²+(y-3)²=5²
the center is the point (2,3) and the radius is r=5 units
Part C)
using a graph tool
see the attached figure
x²+y²-4x-6y-12=0
Group terms that contain the same variable, and move the constant to the opposite side of the equation
(x²-4x)+(y²-6y)=12
Complete the square twice. Remember to balance the equation by adding the same constants to each side
(x²-4x+4)+(y²-6y+9)=12+4+9
Rewrite as perfect squares
(x-2)²+(y-3)²=25
the answer part A) is
(x-2)²+(y-3)²=5²-----> this is the standard form of the equation of a circle
Part B) (x-2)²+(y-3)²=5²
the center is the point (2,3) and the radius is r=5 units
Part C)
using a graph tool
see the attached figure
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