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# Proof: Radius is perpendicular to tangent line

CCSS.Math:

## Video transcript

so what we have here is a circle with the center at point O and then we have a tangent line to the circle and let me actually label this line let's call this line L and we see at Point a is the point that the tangent line intersects with the circle and then we've drawn a radius from the center of the circle to point a now what we want to do in this video is prove to ourselves that this radius and that this tangent line intersect at a right angle we want to prove to ourselves that they intersect at a right angle and the first step to doing that is we're going to feel good we're going to prove to ourselves that point a is the closest point on line L to the center of our circle so I want to prove prove that point a is closest closest point on L to point O two point O and I encourage you to pause the video and see if you can if you can prove that to yourself well to think about that just think about any other point on line L pick any other arbitrary point on line L it could be this point right over here it could be this point right over here it could be this point right over here and you immediately see that it sits outside of the circle and if it's sitting outside of the circle I'll pick this point here just so it'll become a little bit clearer on our diagram if it's sitting outside of the circle in order to get from point O to this point and we call this point B right over here you have to you have to go to the length of the radius you have to go the length of the radius and then you have to go some more so this length the length of segment OB is clearly going to be longer than the length of the radius because you have to go to the radius to get to this to get to the circle itself and then you have to go a little bit further for any point that sits outside of the circle so point a is the only point by definition this is a tangent line it's the only point that sits on the circle every other point on line L sits outside of the circle so it's going to be further so point a hopefully this makes you feel good could you pick any other point is going to sit outside of the circle you have to go to the radius and then some so hopefully this makes you feel good that point a is the closest point on L to the center of the circle now we're not done yet we now have to convince ourselves that if we have a point and a line that the segment connecting the point to the closest point on the line to to that original point then that's going to be perpendicular to the line so let me give ourselves some space here we want to prove we want to prove that if we that the segment segment connecting connecting a point off the line off the line and closest point on the line closest point on the line is perpendicular is perpendicular to the line so what we want to do is we want to say hey if we have some line here if we have some line here L and if you were to take a point off the line so let's say so you wanted a point off the line so let's say that that's this point right over here point O and you want the segment connecting the point off the line to the closest point on the line so the closest point on the line so let's say that this is the closest point on the line we want to feel good that this segment connecting them so let me do this in a new color that the segment connecting them is going to be perpendicular to the line that it goes just straight down like that that is going to be perpendicular and I'm going to prove this by contradiction I'm going to assume that it's not perpendicular so assume assume that the segment segment connecting connecting this is kind of wordy a point off line and closest point to the line closest point to line line is not is not perpendicular to the line so how can I visualize that well I could draw my line right over here so that's line L and let's say I have my point O right over here point O and let's say the closest point on line L 2.0 let's say that it's not so let's say it's over here that if I were to connect these two points that it's not perpendicular to line L so this is the closest point let's call this point a and let's say that the segment connecting these two is not perpendicular to the line so let me get so let's assume this is not perpendicular so this angle this angle this angle right here is not is not 90 degrees so if we assume that the reason why this is going to be a proof by contradiction is I can show that is this is not 90 degrees that I can always find a point that is going to be closer another point on line L that is going to be closer to point O which contradicts the fact that this was supposed to be the closest point a was supposed to be the closest point on line L toe and how do I always find a closer point well I construct a right triangle I can construct a right triangle just like this I can construct a right triangle like that and we see that this distance let's call this distance right over here a unless we could call the base of this triangle B let me just in a different color so a B if that's the base of the right triangle then the hypotenuse is the distance from o to a we could call that C we know from the Pythagorean theorem that a squared plus B squared plus B squared let me give this in a plus B squared is going to be equal to C squared is going to be equal to C squared and so B squared if we have a non degenerate triangle right over here this is going to be some positive value over here and so a is going to be less than C so this gets us to the conclusion because if this is some positive value here and a and C are positive everything at a positive distances and this tells us that a has to be less than C that a non hypotenuse side of a right triangle of a non degenerate right triangle assuming it has some area is going to be shorter than the hypotenuse the hypotenuse is the longest side so a is going to be less than C which would tell us if a is less than C then we found another point let's call this point I've used a lot of letters here let's point let's call that point D D is going to be closer D is closer D D is closer so we've just set up a contradiction we assumed a was the closest point on line L to 0.0 but we assumed that the segment connecting them is not at a 90 degree angle if it's not at a 90 degree angle then we can drop a perpendicular and find a closer point which fot which is a contradiction to the fact that a was supposed to be the closer point so this leads to a contradiction contradiction contra contradiction because you can actually find this is not the closest point you can always find a closer point so therefore the segment connecting a point off the line to the closest point to the line must be perpendicular must be perpendicular so the segment connecting a point off the line to a closest point on the line that must be perpendicular to the line must be perpendicular to the line and just like that we hope we hopefully feel good about the idea that if you have a radius and the point at which it intersects a tangent line to the circle that that forms a 90 degree angle the radius and the tangent line