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How do you find the center of the circle given an equation?
A circle with center (xo, yo) and radius r has an equation of the form: (x - xo)2 + (y - yo)2 = r2 which can be rearranged to: x2 - 2xox + xo2 + y2 - 2yoy + yo2 = r2 ⇒ x2 - 2xox +y2 - 2yoy + xo2 + yo2 - r2 = 0 or x2 - 2xox +y2 - 2yoy = r2 - xo2 - yo2 So depending how your equation is given, the center can be read directly from the equation (first form above) or is -1/2 times the coefficients of the x and y terms. eg: What is the center of the circle with equation x2 - 4x + y2 + 6y + 12 = 0 The coefficients of the x and y terms are -4 and +6, so the center is -1/2 times these, namely; (2, -3). Fully rearranging the circle's equation: x2 - 4x + y2 + 6y + 12 = 0 ⇒ (x - 2)2 - 4 + (y + 3)2 - 9 + 12 = 0 [completing the squares] ⇒ (x - 2)2 + (y + 3)2 - 1 = 0 ⇒ (x - 2)2 + (y + 3)2 = 1 ⇒ circle has centre (2, -3) and radius 1.
How do you solve word equation C6H14 plus O2----CO2 plus H20?
To balance this equation, you must ensure the number of each type of atom is the same on both sides by adjusting coefficients. The balanced equation is C6H14 + 9O2 -> 6CO2 + 7H2O. This means that 1 molecule of C6H14 reacts with 9 molecules of O2 to produce 6 molecules of CO2 and 7 molecules of H2O.
What is the equation of a circle with center 4 -1 and radius of 8?
The general form for the equation of a circle centre (xo, yo) and radius r is: (x - xo)2 + (y - yo)2 = r2 Which can be rearranged into: x2 - 2xox + y2 - 2yoy = r2 - xo2 - yo2 So a circle with centre (4, -1) and radius 8 has equations: (x - 4)2 + (y + 1)2 = 64 x2 - 8x + y2 + 2y = 47
