Find the vertex form of the circle with the equation \(x^2 + y^2 - 6x + 8y + 9 = 0\).
Answer
Finally, the vertex form of the circle is \((x-h)^2 + (y-k)^2 = r^2\), where (h,k) is the center of the circle, and r is the radius. So, the vertex form of the circle is \((x-3)^2 + (y+4)^2 = 4\), where the center is at (3,-4) and the radius is 2.
Step 1 ：First, we will group the x's and y's terms together, the equation will become \(x^2 - 6x + y^2 + 8y = -9\).
Step 2 ：Next, we will complete the square for both x's and y's terms. For x's, we take half of -6, square it, and add it to both sides. For y's, we take half of 8, square it, and also add it to both sides. After doing this, the equation becomes \((x-3)^2 + (y+4)^2 = 4\).
Step 3 ：Finally, the vertex form of the circle is \((x-h)^2 + (y-k)^2 = r^2\), where (h,k) is the center of the circle, and r is the radius. So, the vertex form of the circle is \((x-3)^2 + (y+4)^2 = 4\), where the center is at (3,-4) and the radius is 2.
