What is the center of a circle whose equation is x^2+y^2-12x-2y+12=0
2 answers:
Answer:
centre = (6, 1 )
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Given
x² + y² - 12x - 2y + 12 = 0
Collect the x and y terms together and subtract 12 from both sides
x² - 12x + y² - 2y = - 12
To obtain the equation in standard form use completing the square
add ( half the coefficient of the x/ y term )² to both sides
x² + 2(- 6)x + 36 + y² + 2(- 1)y + 1 = - 12 + 36 + 1
(x - 6)² + (y - 1)² = 25 ← in standard form
with centre = (6, 1 ) and radius = = 5
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Answer:
x = -
Step-by-step explanation:
To find f(g(x)) substitute x = g(x) into f(x), that is
f(g(x))
= f(x + 1)
= 2(x + 1)² ← expand using FOIL
= 2(x² + 2x + 1) ← distribute
= 2x² + 4x + 2
To find g(f(x)) substitute x = f(x) into g(x), that is
g(f(x))
= g(2x²)
= 2x² + 1
----------------------------------------------------------
Equating gives
2x² + 4x + 2 = 2x² + 1 ( subtract 2x² + 1 from both sides )
4x + 1 = 0 ( subtract 1 from both sides )
4x = - 1 ( divide both sides by 4 )
x = -
Answer:
x = 4 or x = -3/2
Step-by-step explanation:
To simplify and solve this, let's use the quadratic formula.
So the format of a quadratic equation is supposed look like this:
Let's find a, b, c:
- a is the coefficient of x^2
- b is the coefficient of x
- c is the constant term
Now let's solve the equation.
Let's solve your equation step-by-step.
2x^2−5x−12=0
For this equation: a=2, b=-5, c=-12
2x^2+−5x+−12=0
Step 1: Use quadratic formula with a=2, b=-5, c=-12.
or
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<u>Answer:</u>
x = 4 or x = -3/2