The circle \( t \) has equation \( x^{2}+y^{2}=25 \). The point \( D(-4,3) \) is on the circle \( t \). Find the equation of \( k \), the tangent to the circle \( t \) at the point \( D \). Give your answer in the form \( a x+b y+c=0 \), where \( a, b c \in \mathbb{Z} \).

To find the equation of the tangent line \( k \) to the circle \( t \) at the point \( D(-4, 3) \), we can follow these steps: 1. **Identify the center and radius of the circle**: The equation of the circle is given by \( x^2 + y^2 = 25 \). This can be rewritten in standard form as: \[ (x - 0)^2 + (y - 0)^2 = 5^2 \] From this, we see that the center of the circle is \( C(0, 0) \) and the radius \( r = 5 \). 2. **Find the slope of the radius at point \( D \)**: The slope of the line connecting the center \( C(0, 0) \) to the point \( D(-4, 3) \) is calculated as follows: \[ \text{slope of } CD = \frac{y_D - y_C}{x_D - x_C} = \frac{3 - 0}{-4 - 0} = \frac{3}{-4} = -\frac{3}{4} \] 3. **Determine the slope of the tangent line**: The tangent line at point \( D \) will be perpendicular to the radius \( CD \). The slope of the tangent line \( k \) is the negative reciprocal of the slope of the radius: \[ \text{slope of } k = -\frac{1}{\left(-\frac{3}{4}\right)} = \frac{4}{3} \] 4. **Use point-slope form to find the equation of the tangent line**: The point-slope form of a line is given by: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is the point \( D(-4, 3) \) and \( m \) is the slope of the tangent line. Substituting the values: \[ y - 3 = \frac{4}{3}(x + 4) \] 5. **Rearranging to standard form**: First, we multiply both sides by 3 to eliminate the fraction: \[ 3(y - 3) = 4(x + 4) \] Expanding both sides: \[ 3y - 9 = 4x + 16 \] Rearranging gives: \[ 4x - 3y + 25 = 0 \] Thus, the equation of the tangent line \( k \) in the form \( ax + by + c = 0 \) is: \[ \boxed{4x - 3y + 25 = 0} \]