To find the equation of the tangent to the circle at the point (1,3), we need to find the slope of the tangent and its point of tangency. The slope of the tangent is -1/(y+1) and the point of tangency is (1,3). Substituting into the point-slope form of the equation of a line, we get y = -x + 4 as the equation of the tangent.
The equation of a circle with center (h,k) and radius r is given by the equation (x-h)^2 + (y-k)^2 = r^2. We can rewrite the equation of the given circle as (x+2)^2 + (y+1)^2 = 25.
To find the equation of the tangent to the circle at the point (1,3), we need to find the slope of the tangent and its point of tangency.
First, we find the slope of the tangent by taking the derivative of the equation of the circle with respect to x. Differentiating implicitly, we get (x+2) + (y+1) * dy/dx = 0. Solving for dy/dx, we find that the slope of the tangent is -1/(y+1).
Next, substituting the coordinates of the point (1,3) into the equation of the circle, we get (1+2)^2 + (3+1)^2 = 25. Simplifying, we find that the point (1,3) lies on the circle and has a radius of 5.
Using the point-slope form of the equation of a line, we can substitute the slope of the tangent (-1/(y+1)) and the coordinates of the point (1,3) into the equation y - 3 = -1/(y+1)(x - 1). Simplifying, we get the equation of the tangent as y = -x + 4.