find the center (h,k) and radius r of the circle with the given equation:
x^2 + y^2 + 6x - 8y = 56
This can be done by completing the square, namely we group the x² with the 6x term and add a constant to make it a complete square. The same goes for y.
x²+6x+ y²-8y = 56
(x+3)² -9 + (y-4)² -16 = 56
(x+3)² + (y-4)² = 56 + 9 + 16 = 81 = 9²
Thus the equation of the circle is
(x+3)² + (y-4)² = 9²
meaning that the centre is at (-3,4) and the radius is 9.
To find the center and radius of a circle given its equation, we need to rewrite the equation in the standard form: (x - h)^2 + (y - k)^2 = r^2.
Let's start by completing the square for both the x and y terms of the equation:
x^2 + 6x + y^2 - 8y = 56
For the x terms:
Step 1: Take half of the coefficient of x, which is 6, and square it: (6/2)^2 = 9.
Step 2: Add this value to both sides of the equation: x^2 + 6x + 9 + y^2 - 8y = 56 + 9.
For the y terms:
Step 1: Take half of the coefficient of y, which is -8, and square it: (-8/2)^2 = 16.
Step 2: Add this value to both sides of the equation: x^2 + 6x + 9 + y^2 - 8y + 16 = 56 + 9 + 16.
Now, let's simplify the equation:
(x^2 + 6x + 9) + (y^2 - 8y + 16) = 81.
To further simplify, we can rewrite it as:
(x + 3)^2 + (y - 4)^2 = 81.
Now the equation is 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.
Comparing this to the standard form, we can determine that the center of the circle is (-3, 4) and the radius is the square root of 81, which is 9.
Therefore, the center (h, k) is (-3, 4), and the radius r is 9.