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=(4-0)/(3-0)(x-0)`

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

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