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 start by determining the center and radius of the circle described by the equation \( x^2 + y^2 = 25 \). The center is at the origin \( (0, 0) \) and the radius is 5. The slope of the radius to point \( D(-4, 3) \) can be computed using the formula for the slope between two points: \[ \text{slope of radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{-4 - 0} = -\frac{3}{4}. \] The slope of the tangent line \( k \) is the negative reciprocal of the slope of the radius: \[ \text{slope of tangent} = \frac{4}{3}. \] Now we can use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1), \] where \( (x_1, y_1) \) is the point \( D(-4, 3) \) and \( m = \frac{4}{3} \): \[ y - 3 = \frac{4}{3}(x + 4). \] To clear the fractions, we can multiply everything by 3: \[ 3(y - 3) = 4(x + 4). \] Expanding: \[ 3y - 9 = 4x + 16. \] Rearranging gives: \[ 4x - 3y + 25 = 0. \] Thus, the equation of the tangent line \( k \) in the desired form \( ax + by + c = 0 \) is: \[ \boxed{4x - 3y + 25 = 0}. \]