Find the center and radius of the circle with equation x2 + y2 - 6x + 10y + 9 = 0.
center ( ? , ? )
radius ?
To find the center and radius of a circle given its equation, you need to rewrite the equation in the standard form of a circle, which is (x - h)² + (y - k)² = r², where (h, k) represents the coordinates of the center of the circle, and r represents the radius.
Let's use completing the square method to convert the given equation into the standard form:
x² + y² - 6x + 10y + 9 = 0
Rearrange the terms by grouping the x-terms together and the y-terms together:
(x² - 6x) + (y² + 10y) + 9 = 0
To complete the square for the x-terms, take half of the coefficient of x (-6/2 = -3), square it (-3)² = 9, and add it inside the parentheses. Do the same for the y-terms:
(x² - 6x + 9) + (y² + 10y + 25) + 9 = 9 + 9 + 25
Simplify:
(x - 3)² + (y + 5)² + 9 = 43
To isolate the squared terms on the left side, subtract 9 from both sides:
(x - 3)² + (y + 5)² = 34
Now we have the equation in the standard form. Comparing it to the standard form, we can determine the center and radius.
The center of the circle is located at the point (h, k), which corresponds to the values inside the parentheses. So the center is (3, -5).
The radius (r) is the square root of the value on the right side of the equation, so the radius is √34.
Therefore, the center of the circle is (3, -5), and the radius is √34.
To find the center and radius of the circle, we first need to rewrite the equation in the standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Given equation: x^2 + y^2 - 6x + 10y + 9 = 0
To complete the square for the x terms, we need to add and subtract the square of half the coefficient of x from both sides of the equation:
x^2 - 6x = -y^2 - 10y - 9
(x - 3)^2 - 9 = -y^2 - 10y - 9
To complete the square for the y terms, we need to add and subtract the square of half the coefficient of y from both sides of the equation:
(x - 3)^2 - 9 + 25 = -y^2 - 10y - 9 + 25
(x - 3)^2 + 16 = -y^2 - 10y + 16
Rearranging the terms:
(x - 3)^2 + y^2 + 10y = 0
To complete the square for the y terms, we need to add and subtract the square of half the coefficient of y from both sides of the equation:
(x - 3)^2 + y^2 + 10y + 25 = 0 + 25
(x - 3)^2 + (y + 5)^2 = 25
Comparing this with the standard form, we can see that the center is at (3, -5) and the radius is √25 = 5.
Therefore, the center of the circle is (3, -5) and the radius is 5.
just complete the squares for x and y:
x^2-6x + y^2+10y = -9
x^2-6x+9 + y^2+10y+25 = -9+9+25
(x-3)^2 + (y+5)^2 = 25
I assume you can take it from there, eh?