Finding the Equation of a Line Parallel to a Given Line and Containing the Diameter of a Circle

In this article, we will solve for the equation of the line that is parallel to the line 3x 5y - 4 0 and contains the diameter of the circle given by the equation x^2 y^2 2x - 4y - 4 0. We will break down the process into manageable steps and derive the final equation.

Step 1: Determine the Center of the Circle

We start by converting the circle's equation into the standard form of a circle, which is (x - h)^2 (y - k)^2 r^2, where (h, k) is the center and r is the radius.

The given circle equation is:

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

To convert this to the standard form, we complete the square for both x and y terms:

x^2 2x

Add and subtract (2/2)^2 1: x^2 2x 1 - 1 (x 1)^2 - 1

y^2 - 4y

Add and subtract (-4/2)^2 4: y^2 - 4y 4 - 4 (y - 2)^2 - 4

Now substitute these back into the original equation:

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

Simplify:

(x 1)^2 (y - 2)^2 - 9 0

So, the equation of the circle in standard form is:

(x 1)^2 (y - 2)^2 9

From this, we can see that the center of the circle is (-1, 2).

Step 2: Determine the Slope of the Line

We are given the line 3x 5y - 4 0. To find its slope, we convert it to the slope-intercept form y mx b.

Rearrange:

3x 5y - 4 0

Add -3x 4 to both sides:

5y -3x 4

Divide by 5:

y -frac{3}{5}x frac{4}{5}

The slope of the given line is -frac{3}{5}.

Step 3: Find the Equation of the Line

Since the line we need is parallel to the given line, it will have the same slope -frac{3}{5}. This line also passes through the center of the circle, which is (-1, 2).

Using the slope-point form of a line, y - y1 m(x - x1), where (x1, y1) (-1, 2) and m -frac{3}{5}, we get:

y - 2 -frac{3}{5}(x 1)

Simplify:

y - 2 -frac{3}{5}x - frac{3}{5}

Multiply through by 5 to clear the fraction:

5y - 10 -3x - 3

Bring all terms to one side of the equation:

3x 5y 7

Conclusion

The equation of the line that is parallel to 3x 5y - 4 0 and passes through the center of the circle (-1, 2) is:

3x 5y 7