Question

Consider the circle of radius 5 centered at (0,0), how do you find an equation of the line tangent to the circle at the point (3,4)?

Answer 1

$\text{We know that the tangent at any point on a circle is perpendicular}\\ \text{to the radius at that point.}\\ \\ \text{so first we find the equation of the line joining (0,0) and (3,4)}\\ \text{and then using the slope, we find the line perperndicular to it at (3,4)}\\ \\ \text{the equation of the line passing through (0,0) and (3,4) is}\\ \\ y-0=\frac{4-0}{3-0}(x-0)$

$\Rightarrow y=\frac{4}{3}x\\ \\ \text{so the slope of radius is }m=\frac{4}{3}.\\ \\ \text{we know that the product of the perpendicular lines is }-1. \\ \\ \text{so the slope of the perpendicular line would be}=-\frac{1}{4/3}=-\frac{3}{4}\\ \\ \text{So the equation of the tangent line has slope }-\frac{3}{4} \text{ and}\\ \text{passing through (3,4). so equation of tangent line is}$

$y-4=-\frac{3}{4}(x-3)\\ \\ \Rightarrow y-4=-\frac{3}{4}x+\frac{9}{4}\\ \\ \Rightarrow y=-\frac{3}{4}x+\frac{9}{4}+4\\ \\ \Rightarrow y=-\frac{3}{4}x+\frac{9+16}{4}\\ \\ \text{so the equation of tangent line is:} y=-\frac{3}{4}x+\frac{25}{4}$