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the circle x squared plus y squared minus 8x equals zero and the hyperbola x squared over nine minus y squared over four equal one intersect at the points a and B equation of a common tangent with positive slope so of a common tangent with positive slope to the circle as well as to the hyperbola is so let's just visualize what they're asking first and this is going to take us multiple videos I have a feeling but let's just visualize it just so we can header get our head around the problem so this circle let me complete the square in terms of X so this circle as they wrote it is x squared minus 8x and then we have a plus y squared I left some space here so that we can complete the square is equal to 0 and then let me add half of this 8 term squared to both sides so half of negative 8 is negative 4 negative 4 squared is 16 so add 16 to both sides and that allowed me to turn the X term into a complete into a square perfect square this is the same thing as X minus 4 squared and then we have plus y squared plus y squared is equal to 16 so this is a circle this right here is a circle with Center at X is equal to 0 sorry it x equals 4 at x equals 4 y is equal to 0 and it has a radius it has a radius of 4 as well so let me graph let me graph this circle here so let me draw the horizontal axis my x-axis let me draw the y-axis that is my Y axis over here and let me draw its Center so 1 2 3 4 4 comma 0 that's its Center and has a radius of 4 so it's going to come out it's going to look something it's going to look something I could draw better circle than that it's going to look something like that's the top half and then the bottom half is going to look something like that so that's our circle now let's think about the hyperbola let's think about the hyperbola so if we just look at it the x squared term is positive so it's going to be a hyperbola that opens to the right and the left and we do this a bunch in the conic sections videos if you want a review of that and we could just figure out where where it intersects the x axis so then Y when y is equal to 0 we have x squared over 9 must be equal to 1 or X would be plus or minus 3 so the hyperbola is going to look like this so this is at plus 3 comma 0 the hyperbola will open up like that will open up like that and then at 1 2 3 negative 3 comma 0 the hyperbola is going to open up to the left and so in the problem when they describe the points a and B they're probably talking about that point a and that point B now let's think about what this question is asking us equation of a common tangent with positive slope so that's to have a positive slope to the circle as well as to the hyperbola a common tangent so let's just think about this a little bit so it's going to have a positive slope so it's going to have to so it won't it won't be tangent to the circle anywhere where the circle has a negative slope so it can't be tangent it can't be tangent over here can't be tangent over there and then we could say well what if it was tended to the circle over here what would happen well it would wouldn't be able to be tangent it wouldn't be able to be tangent to the hyperbola so it has to be the tent it has to be tangent to the circle someplace someplace in this blue region right over here and then how can it be tangent to the hyperbola might be tempting to say that it would be tangent to the hyperbola in this way in this way somehow but what you need to realize is the hyperbola is asymptoting towards some line and we could figure out we could figure out what that line is it's asymptoting towards some line so let me draw that line it's asymptoting and so it's always it's always going to have the hyperbola is always going to have a higher slope than that line is very slightly higher slope is slowly approaching that line so if you go out here the hyperbola is going to have a higher slope than that then the asymptote line and so if you had to be tangent to it you would have to have a higher slope you would have to have a higher slope and anything that's coming from this part of the circle towards anything out here on the hyperbolas going to have to have a lower slope than the tangent line right because it's going to have to catch up the tangent line is going to have to catch up to whatever - whatever let me draw it again so whatever we draw over here I want to make this clear the hyperbola the hyperbola as you go out here actually this whole period this whole area this whole part of the hyperbola is going to have a higher slope than what it is asymptoting towards that's what allows it to get closer and closer to that line so any tangent is going to have to have a higher slope out here anything tangent to the hyperbola is going to have to have a higher slope than this actual line it's going to have to have a slightly higher slope so if we take something out here and we try to draw a tangent from this part of the circle out there it wouldn't work because this this tangent by definition will have to in order to meet the hyperbola is going to have to have a lower slope than this asymptote so this can't be tangent to that part of the hyperbola so what else can we do well the only other part of the hyperbola that we might be able to work something out is is this part of the hyperbola right over here so if we find if we find a line that's tangent maybe there and there on the hyperbola then we might have found our common tangent with positive slope so let me draw that so our common tangent with positive slope let me do that in I'll do that in pink our common tangent with positive slope could look like that so now that we have the visualization down in the next video let's try to figure out let's forget what that what that line might look like especially when we constrain it by it when we constrain it to having to beat hand it to this circle and having to be tangent to this hyperbola