Q.Find the equation of the tangent to the circle $x^2+y^2-6x-4y=0$ at the point $(6,4)$ on it.

A tangent to a circle is a line that meets the circle at exactly one point (technically, two coincident points), called the point of contact. Because the radius drawn to the point of contact is always perpendicular to the tangent there, the tangent's slope is the negative reciprocal of the radius's slope — and carrying this through algebraically for the standard circle $x^2+y^2=r^2$ gives the tangent at a point $(x_1,y_1)$ on the circle as

For the general-form circle $x^2+y^2+2gx+2fy+c=0$, the same idea gives $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$ — rememberable as the substitution rule "$x^2\to xx_1$, $2x\to x+x_1$, $y^2\to yy_1$, $2y\to y+y_1$" applied to the circle's own equation. In parametric form, the tangent at $(r\cos\theta_1,r\sin\theta_1)$ is $x\cos\theta_1+y\sin\theta_1=r$. These formulas apply whenever the point of contact is already known to lie on the circle; when a line is only claimed to be tangent (its point of contact not given), tangency is …