Determine if the following system of equations has no solutions, infinitely many
1 answer:
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Remember that the equation of a circle is:
Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.
Complete the square. We must do this for x² - 6x and y² - 10y separately.
x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9
Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.
Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.
y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.
Factor y² - 10y + 25 to get (y - 5)²
Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)
Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.
Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.
Complete the square. We must do this for x² - 6x and y² - 10y separately.
x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9
Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.
Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.
y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.
Factor y² - 10y + 25 to get (y - 5)²
Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)
Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.
