Prove that the line lx+my+n=0 toches the circle `(x-a)^(2)+(y-b)^(2)=r^(2)` if `(al+bm+n)^(2)=r^(2)(l^(2)+m^(2))`

Answer

To prove that the line \( lx + my + n = 0 \) touches the circle \( (x-a)^2 + (y-b)^2 = r^2 \) if \( (al + bm + n)^2 = r^2(l^2 + m^2) \), we will follow these steps:

### Step 1: Identify the Circle and the Line
We have the equation of the circle:
\[ (x - a)^2 + (y - b)^2 = r^2 \]
This circle has its center at \( (a, b) \) and radius \( r \).

The line is given by:
\[ lx + my + n = 0 \]

### Step 2: Calculate the Perpendicular Distance from the Center of the Circle to the Line
The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
In our case, the coefficients are:
- \( A = l \)
- \( B = m \)
- \( C = n \)

The center of the circle is \( (a, b) \). Thus, the distance \( d \) from the center \( (a, b) \) to the line \( lx + my + n = 0 \) is:
\[ d = \frac{|la + mb + n|}{\sqrt{l^2 + m^2}} \]

### Step 3: Set the Distance Equal to the Radius
Since the line is a tangent to the circle, the distance from the center of the circle to the line must equal the radius \( r \):
\[ \frac{|la + mb + n|}{\sqrt{l^2 + m^2}} = r \]

### Step 4: Square Both Sides
To eliminate the square root and the absolute value, we square both sides:
\[ \left( \frac{la + mb + n}{\sqrt{l^2 + m^2}} \right)^2 = r^2 \]
This simplifies to:
\[ \frac{(la + mb + n)^2}{l^2 + m^2} = r^2 \]

### Step 5: Rearranging the Equation
Multiplying both sides by \( l^2 + m^2 \) gives:
\[ (la + mb + n)^2 = r^2(l^2 + m^2) \]

### Conclusion
Thus, we have shown that the line \( lx + my + n = 0 \) touches the circle \( (x-a)^2 + (y-b)^2 = r^2 \) if and only if:
\[ (al + bm + n)^2 = r^2(l^2 + m^2) \]