Show that the line `5x + 12y - 4 = 0 `
touches the circle
`x^(2)+ y^(2) -6x + 4y + 12 = 0`

Answer

To show that the line \(5x + 12y - 4 = 0\) touches the circle given by the equation \(x^2 + y^2 - 6x + 4y + 12 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 6x + 4y + 12 = 0 \]
We can rearrange this equation to find the center and radius of the circle. We will complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 6x \rightarrow (x - 3)^2 - 9 \]

2. For \(y\):
\[ y^2 + 4y \rightarrow (y + 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 + 12 = 0 \]
Simplifying this:
\[ (x - 3)^2 + (y + 2)^2 - 1 = 0 \]
Thus, we have:
\[ (x - 3)^2 + (y + 2)^2 = 1 \]
This shows that the center of the circle is \((3, -2)\) and the radius \(R = 1\).

### Step 2: Find the distance from the center of the circle to the line
The line is given by:
\[ 5x + 12y - 4 = 0 \]
We can use the formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 5\), \(B = 12\), \(C = -4\), and the center of the circle is \((3, -2)\).

Substituting these values into the distance formula:
\[ d = \frac{|5(3) + 12(-2) - 4|}{\sqrt{5^2 + 12^2}} \]

Calculating the numerator:
\[ = |15 - 24 - 4| = |-13| = 13 \]

Calculating the denominator:
\[ = \sqrt{25 + 144} = \sqrt{169} = 13 \]

Thus, the distance \(d\) is:
\[ d = \frac{13}{13} = 1 \]

### Step 3: Compare the distance with the radius
We found that the distance from the center of the circle to the line is \(1\), and the radius of the circle is also \(1\). Since the distance equals the radius, we conclude that the line touches the circle.

### Conclusion
The line \(5x + 12y - 4 = 0\) touches the circle \(x^2 + y^2 - 6x + 4y + 12 = 0\).

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