Q.If the length of the tangent from $(2, 5)$ to the circle $x^2 + y^2 - 5x + 4y + k = 0$ is $\sqrt{37}$ then find k.

Use the length-of-tangent formula $\sqrt{S_1}$ where $S_1$ is the circle's expression evaluated at the external point.

For a circle $x^2+y^2+2gx+2fy+c=0$, the length of the tangent from a point $(x_1,y_1)$ outside it is $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.

Here the circle is $x^2+y^2-5x+4y+k=0$, so $2g=-5$, $2f=4$, $c=k$. The point is $(2,5)$.

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