Find the equation of the tangent line at the given point on the following curve. \[ x^{2}+y^{2}=26,(1,5) \]

To find the equation of the tangent line at the point (1,5) on the curve \( x^{2}+y^{2}=26 \), we first need to find the derivative of the curve implicitly. Differentiating both sides gives: \[ 2x + 2y\frac{dy}{dx} = 0. \] Rearranging this, we find: \[ \frac{dy}{dx} = -\frac{x}{y}. \] Now, substituting the point (1,5) into the derivative: \[ \frac{dy}{dx} = -\frac{1}{5}. \] The slope of the tangent line at the point (1,5) is \(-\frac{1}{5}\). We can use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1), \] where \( m \) is the slope and \((x_1,y_1)\) is the point on the line. Substituting in the values: \[ y - 5 = -\frac{1}{5}(x - 1). \] Multiplying through by 5 to eliminate the fraction gives: \[ 5(y - 5) = -(x - 1) \] or \[ 5y - 25 = -x + 1. \] Rearranging gives us: \[ x + 5y = 26. \] So, the equation of the tangent line at the point (1,5) is: \[ x + 5y = 26. \]