How do you find the center and radius for #x^2 + y^2 - 6x - 4y - 12 = 0#?
Use the quadratic 'Complete the Square' method
#x^2 - 6x +y^2 - 4y = 12
Then take 1/2 of the 'b' term for both quadratic expressions, square those values and add them to both sides.
#x^2 -6x + 9 + y^2 - 4y + 4 = 12 + 9 + 4
(x - 3)^2 + (y -2)^2 = 25
Circle centered at (3,2) with radius = 5
An equation in the form:
is the standard form for the equation of a circle with center
Lets try to convert the given equation:
into the standard form for the equation of a circle.
Complete the square for each of
Write the left side as the sum of two squared binomials
and simplify the result on the right side
Express the right side as a square.
...the equation for a circle with center