prove that the line `3x+4y+7=0` touches the circle `x^2+y^2-4x-6y-12=0` also find the point of contact

Answer

To prove that the line \(3x + 4y + 7 = 0\) touches the circle \(x^2 + y^2 - 4x - 6y - 12 = 0\) and to find the point of contact, we will follow these steps:

### Step 1: Find the center and radius of the circle

The general equation of a circle is given by:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From the given circle equation \(x^2 + y^2 - 4x - 6y - 12 = 0\), we can identify:

- \(2g = -4 \Rightarrow g = -2\)
- \(2f = -6 \Rightarrow f = -3\)
- \(c = -12\)

The center of the circle \((h, k)\) is given by:

\[ (h, k) = (-g, -f) = (2, 3) \]

The radius \(r\) of the circle is given by:

\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5 \]

### Step 2: Find the length of the perpendicular from the center to the line

The formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the line \(3x + 4y + 7 = 0\), we have:

- \(A = 3\)
- \(B = 4\)
- \(C = 7\)

Using the center of the circle \((2, 3)\):

\[ d = \frac{|3(2) + 4(3) + 7|}{\sqrt{3^2 + 4^2}} = \frac{|6 + 12 + 7|}{\sqrt{9 + 16}} = \frac{|25|}{\sqrt{25}} = \frac{25}{5} = 5 \]

### Step 3: Compare the distance with the radius

Since the distance \(d = 5\) is equal to the radius \(r = 5\), we conclude that the line touches the circle.

### Step 4: Find the point of contact

To find the point of contact, we need to find the intersection of the line and the radius from the center of the circle to the line.

1. The slope of the line \(3x + 4y + 7 = 0\) can be rewritten in slope-intercept form:

\[ 4y = -3x - 7 \Rightarrow y = -\frac{3}{4}x - \frac{7}{4} \]

The slope of this line is \(-\frac{3}{4}\).

2. The slope of the radius (which is perpendicular to the line) is the negative reciprocal:

\[ m = \frac{4}{3} \]

3. The equation of the radius line passing through the center \((2, 3)\) is:

\[ y - 3 = \frac{4}{3}(x - 2) \]

Rearranging gives:

\[ y = \frac{4}{3}x + \left(3 - \frac{8}{3}\right) = \frac{4}{3}x + \frac{1}{3} \]

4. Now, we have two equations to solve:

- Line: \(3x + 4y + 7 = 0\)
- Radius: \(4x - 3y + 1 = 0\)

5. To find the point of intersection, we can multiply the first equation by 3 and the second by 4 to eliminate \(y\):

\[ 9x + 12y + 21 = 0 \quad \text{(1)} \]
\[ 16x - 12y + 4 = 0 \quad \text{(2)} \]

Adding (1) and (2):

\[ 25x + 25 = 0 \Rightarrow x = -1 \]

Substituting \(x = -1\) back into the line equation:

\[ 3(-1) + 4y + 7 = 0 \Rightarrow -3 + 4y + 7 = 0 \Rightarrow 4y + 4 = 0 \Rightarrow y = -1 \]

Thus, the point of contact is:

\[ (-1, -1) \]

### Final Result

The line \(3x + 4y + 7 = 0\) touches the circle \(x^2 + y^2 - 4x - 6y - 12 = 0\) at the point \((-1, -1)\).

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