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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.
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
How do you solve word equation C6H14 plus O2----CO2 plus H20?
You have to think a bit 'obtusely/backwardly'. C6H14 + O2 = CO2 + H2O We note that the alkane has '6' carbons , so we can put a '6' in front of the CO2 Similarly, there are '14' hydrogens , so we put a '7' in front of the 'H2O' . There being already two atoms of hydrogen in each molecule of water , so 7 x 2 = 14 Hence C6H14 + O2 = 6CO2 + 7H2O This is where we think obtusely. We note that there are 6 x 2 = 12 oxygens in CO2 , and 7 x 1 = 7 oxygens in H2O. So on the product side we have 12 + 7 = 19 oxygens. So on the reactant side the 'O2' has to be made up to '19' oxygens. This is done by placing a molar ratio of 19 = 9/2 on the reacting oxygens. Hence C6H14 + 19O2 = 6CO2 + 7H2O This gives us '38' oxygens. So to balance the other three components are multiplied by '2' Hence 2C6H14 + 19O2 = 12CO2 + 14H2O Balanced. You will notice There are 2 x 6 = 12 (C) 2 x 14 = 14 x 2 =28(H) 19 x 2 = (12 x 2)+ (14 x 1) = 38 (O) NB In working out these organic combustion problems. Initially balance the numbers of 'C' & 'H' on the product side. Then look at the number of oxygens on the product side, and working back to the number of oxygens required to react.
