91Ó°ÊÓ

When dealing with the equations of circles, it's important to know how to rewrite them in the standard form. The standard form of a circle is expressed as: \[ (x - h)^2 + (y - k)^2 = r^2 \]. Here, \( (h, k) \) is the center of the circle, and \( r \) is the radius.
For example, we can take the circle's equation from the problem: \( x^2 + y^2 - 6x + 2y - 17 = 0 \). To rewrite this in standard form, we need to complete the square for both the x and y terms. This means we group x terms together and y terms together and then complete the square for each group.
The equation becomes \( x^2 - 6x + y^2 + 2y = 17 \).

Completing the square is a technique used to transform a quadratic equation into a perfect square trinomial, which can make it easier to solve or identify components like the vertex of a parabola or the center of a circle.
For \( x^2 - 6x \), we find the number that completes the square. Take half of the x coefficient (which is -6), square it, and add and subtract this number: \[ x^2 - 6x = (x - 3)^2 - 9 \].
Similarly for \( y^2 + 2y \), take half of 2, square it, and add and subtract this value: \[ y^2 + 2y = (y + 1)^2 - 1 \].
Now, we combine both completed squares in the equation: \[ (x - 3)^2 - 9 + (y + 1)^2 - 1 = 17 \].
Simplify this to get the standard form of the circle: \[ (x - 3)^2 + (y + 1)^2 = 27 \].

The quadratic formula is a straightforward method to find the solution for a quadratic equation of the form \[ ax^2 + bx + c = 0 \]. It is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
Let's use this formula to solve \[ 2y^2 - 8y - 1 = 0 \], which we derived in the exercise. Here, \( a = 2 \, b = -8 \, \text{and}\ c = -1 \): \[ y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{8 \pm \sqrt{64 + 8}}{4} = \frac{8 \pm \sqrt{72}}{4} = \frac{8 \pm 6 \sqrt{2}}{4} = 2 \pm \frac{3 \sqrt{2}}{2} \. \]
Therefore, the values of y are \[ y = 2 + \frac{3 \sqrt{2}}{2} \] and \[ y = 2 - \frac{3 \sqrt{2}}{2} \].
Using \( x = y - 2 \, \text{we find the x-values}\: \) \ \[ x_1 = \frac{3 \sqrt{2}}{2} \] and \[ x_2 = -\frac{3 \sqrt{2}}{2} \. \]
Hence, the points of intersection are: \ \ \[ \textrm{(\frac{3 \sqrt{2}}{2}, 2 + \frac{3 \sqrt{2}}{2})} \] and \ \[ \textrm{(-\frac{3 \sqrt{2}}{2}, 2 - \frac{3 \sqrt{2}}{2})} \. \]