How to Find the Equation of a Line Perpendicular to ( y x - 5 ) and Passing Through the Center of a Circle

To find the equation of the straight line that goes through the center of the given circle and is perpendicular to the line ( y x - 5 ), we need to follow a series of steps involving algebraic manipulation and the application of geometric principles. Let's break down the process step by step.

Step 1: Find the Center of the Circle

The given equation of the circle is:

2x^2 - 2y^2 - 4y - 1  0

First, we simplify this equation by dividing the entire equation by 2:

x^2 - y^2 - 2y - frac{1}{2}  0

Next, we rearrange it to the standard form:

x^2 - y^2 - 2y  frac{1}{2}

Now, we complete the square for the (y) terms:

 y^2 - 2y  (y - 1)^2 - 1

Substituting back into the equation, we get:

x^2 - (y - 1)^2   1  frac{1}{2}

This simplifies to:

x^2 - (y - 1)^2  -frac{1}{2}

Further rearranging the equation, we have:

x^2 - (y - 1)^2   frac{1}{2}  0

From the standard form of the circle ((x - h)^2 (y - k)^2 r^2), we can see that:

x^2 - (y - 1)^2  frac{1}{2}

This tells us that the center of the circle is at the point ((0, -1)).

Step 2: Find the Slope of the Line Perpendicular to ( y x - 5 )

The slope of the line ( y x - 5 ) is 1. The slope of a line perpendicular to this line is the negative reciprocal:

m  -1

Step 3: Use the Point-Slope Form to Find the Equation of the Line

Now that we have the center of the circle ((0, -1)) and the slope (-1), we can use the point-slope form of the equation of a line:

y - y_1  m(x - x_1)

Substituting ( m -1 ), ( x_1 0 ), and ( y_1 -1 ) into the equation, we get:

y - (-1)  -1(x - 0)

This simplifies to:

y   1  -x

Rearranging gives us:

y  -x - 1

Final Answer

The equation of the straight line that goes through the center of the circle and is perpendicular to the line ( y x - 5 ) is:

boxed{y  -x - 1}

Additional Insights

To further understand the problem, let's break down the reasoning involved:

Identify the center of the circle: Given the equation (2x^2 - 2y^2 - 4y - 1 0), we simplify and complete the square to find the center at ((0, -1)).

Determine the slope of the perpendicular line: The slope of the original line is 1, so the slope of the perpendicular line is (-1).

Use the point-slope form: With the center ((0, -1)) and slope (-1), we can plug these values into the point-slope form to derive the final line equation.

Thus, we have successfully identified the equation of the straight line through the center of the circle that is perpendicular to ( y x - 5 ).