Find the equation of the tangent line at the given point on the following curve.
\[
x^{2}+y^{2}=29,(-2,-5)
\]
The equation of the tangent line to the point $(-2,-5)$ is $y=$

Step 1 ：Given the equation of the circle is \(x^{2}+y^{2}=29\) and the point is \((-2,-5)\).

Step 2 ：Rearrange the equation of the circle to get \(y^{2}=29-x^{2}\).

Step 3 ：Take the square root of both sides to get \(y=\sqrt{29-x^{2}}\).

Step 4 ：Find the derivative of y with respect to x to get \(y'=-\frac{x}{\sqrt{29-x^{2}}}\).

Step 5 ：Substitute \(x=-2\) into the derivative to find the slope of the tangent line at the point \((-2,-5)\), which is \(y'(-2)=\frac{2}{5}\).

Step 6 ：Substitute the point \((-2,-5)\) and the slope \(\frac{2}{5}\) into the equation of a line \(y=mx+b\) to find the y-intercept, which is \(b=-4.2\).

Step 7 ：\(\boxed{The equation of the tangent line at the point (-2,-5) is y=\frac{2}{5}x-4.2}\).