okay welcome to this video where we're going to have a look at everything you need to do with circles when it comes to the asp maths curriculum and you can see the topics on the screen that we're going to be having a look at we'll be looking at midpoints and perpendicular bisectors the equation of a circle intersections of straight lines and circles tangents and chord properties and circles with triangles so for this video we're going to have a look at quite a few topics it's going to cover the whole of this chapter and as always i will link this all the videos in the description and also show you the chapters so very quickly before we get started i will show you how that works so when you are on one of these videos hopefully you've seen them before and hopefully you'll have seen this video here where i cover the entire asp curriculum now you'll notice that when you're on one of these videos there are two things that you can do you can click into the description so if we click into that it will bring up all of the timestamps just to the right here that breaks the video up into chapters and because it's broken up into chapters what you can do you can click on the video click into the chapters there just down the bottom left and you can see everything bookmarked for you so that you can just click on any one and it will jump you to that part of the video so hopefully that's useful and helpful don't forget to like the video don't forget to share it with one of your friends and let's get started [Music] okay so when looking at midpoints and perpendicular bisectors this is quite a common type of question so it says the line segment a b is the diameter of a circle center c where a is negative 1 4 and b is 5 2 says the line l passes through c and is perpendicular to the line a b which is the diameter find the equation of line l so we want to find the perpendicular bisector of this diameter and obviously the perpendicular bisector is going to be perpendicular so we have all our stuff to do with perpendicular gradients so if we start by finding the midpoint of these two we might be able to draw a little sketch of this a little bit better so to find the midpoint you take the coordinates and add them together and divide it by two so if we focus first on the x coordinate to get the midpoint of the x coordinate we would do five plus negative one the two x coordinates and divide that by two and the other coordinate would be two plus four and again dividing that by two so for that particular coordinate there five add negative one is four divided by two is two and two plus four is six divided by two is three so we could probably draw an okay sort of sketch and think about what this looks like so a is negative 1 4 so negative 1 4 let's just imagine that's here and then our coordinate in the center is 2 3 so that is to the right obviously and just a little bit lower than the four and then we've got another coordinate which is five two which is to the right again and just a little bit lower so it's going to look something like this now obviously you're not going to be able to draw a perfect sketch of this but you can just have something to help you visualize it so if we imagine it looks something like that although that wasn't very good as my center doesn't look fantastic so you can always get rid of these coordinates and just put them back on so it suits your diagram so let's say it's there there and there so now we have the coordinates of the center so we could also draw in this line a b and let's label that up so that would be a and this would be b and we have our center point there and we are going to find the perpendicular bisector so if we find the or if we draw that in and imagine what it looks like it's going to look something like that and it probably be also worthwhile labeling on that center point 2 three so how do we go about finding a perpendicular line well first of all we want to find the gradient so if we find the gradient of a b well then we can go about doing that to start with so for finding the gradient y two minus y1 so that is going to be the y coordinate of b take away the y coordinate of 1 of a sorry over the x coordinate of b which is 5 take away negative 1 the x coordinate of a so that comes out as negative two over six which is minus a third so our gradient of a b is minus one third which means our perpendicular gradient is going to be our negative reciprocal and that comes out as positive 3. so that is the first thing we need that is the gradient of our perpendicular bisector here so now we can use our line equation so y minus y1 equals mx minus x one and it is going to have to pass through that coordinate through the center which we can see here two and three so we would have y minus the y coordinate of three is equal to m which is three about to draw the bracket let's get rid of that equals three lots of x minus the x coordinate which is two so if we expand that out we get y minus three is equal to three minus 6 and then of course to finish this off we want it just to say y equals so add 3 to both sides there we go and we get y is equal to 3x minus 3 and that is the final line equation and that is the equation of our perpendicular bisector passing through the midpoint of a b in terms of actually the equation of the circle i'm always just going to write that down so i've got x minus a which is the first coordinate or the x coordinate within the center squared add to that y minus b which is obviously the y coordinate of the center squared and that's going to equal the radius squared so if we plug these numbers in that we've been given three and one let's just put them into the equation let's see what we get so there we go we've got x minus three squared add y minus 1 squared and that is equal to the radius squared now i'm going to do this in two different ways so if you've seen this before you might have only used one of the ways so i'm going to show you both but in terms of what we did at gcse maybe at this point we'd have imagined these two coordinates okay so if i imagine the point three one which is the center it does always help if you can draw a little diagram of these but if we imagine three one's the center and then negative two five let's have a just have a little think it probably looks something like this there we go and negative two five is somewhere over here let's have a think negative two fives maybe over there somewhere there we go but in terms of looking at these two points we would find the length of the radius by using pythagoras like we've just talked about and we would look at the base or the or the rise and the run there or the change in x and the change in y and then we just use pythagoras to work out the radius so in terms of the base length it goes from negative two on the left there over to three which is a length of five and going up it goes from one up to five which is a length of four so you can use pythagoras to work out the length of the radius there and we can just do that quite nice and easy we can do four squared plus five squared which is 16 plus 25 which is 41 and then you can just square root that for the radius so r equals the square root of 41. there we go but obviously we want to know what the radius squared is so when it's in a nice little square root like that we can just take away the square root and it'd be 41. so we could finish off our equation here it'll be x minus 3 squared plus y minus 1 squared equals the radius squared the radius is root 41 so 41 so it says the circle c has equation and then we have the equation here and lots of pieces we've got x squares y squared x y's and a number and it all equals zero says find the coordinates of the center of c and the radius of c and then we've got another question here that says find the coordinates of the points where c crosses the y axis so what i've looked at in a sec but this actual equation here has to be rearranged in order to get into the format that we want obviously we know what the format looks like we've got x minus a in bracket squared and then we've got y minus b in bracket squared and that equals r squared so we have to rearrange this we need to get all those numbers on the right hand side and then we need to get these x parts into brackets and if you hopefully recognize this you should recognize that this looks very similar to completed square form and this is where we're going to use a bit of completing the square to get into this format now if we have a look at the pieces that we have in our equation we've got some x pieces here we've got and i'll highlight this in that color there we've got the x squared and we've got a minus 6x now if we just forget about everything else there we group those together we'll have x squared minus 6x okay now if we were to complete the square on that okay just on that part there we would obviously halve the coefficient of x so it'd be x minus three in bracket squared and obviously if you're not sure on completing the square don't forget i've linked that in the description below so check that out first and then obviously we need to get that back to zero because there's no number after this so to get that back to zero thinking about what that would expand to well negative three times negative three would give us positive nine so i'd also have to take away nine and that obviously that expression there would give us x squared minus 6x so i'm just writing that in completed square form we also have the y pieces now if we have a look at the y pieces in there let's just highlight those we've got y squared and a plus 2y now if i put those in we've got so i'll put a plus here as well we've got y squared plus 2y now if i complete the square just on that and let's keep the plus there as well we would have the coefficient of y so i'd get y plus one there we go in bracket squared when we expand that that gives us plus one so we have to take that one away to make it balance to the expression above the y squared plus two y and that would obviously complete the square on that part so so far what we've done we've completed the square there that's given us a little bit below and we completed the square for the x parts there again for that bit below as well then we've still got that minus 15 so i need to bring that minus 15 back in there we go and it all equals zero so add the minus 15 in there we go and that all equals zero now if we have a look at just the numbers that we've got there we've got a minus nine a minus one and a minus fifteen now if i group those all together and i'll probably skip a few steps on the next question here but if i group all the numbers together and i'll put them all at the end let's see what we've got we've got x minus three in bracket squared plus the y plus one in bracket squared and then all of those numbers so we've got minus nine minus one that's minus 10 minus another 15 is minus 25 there we go and that all equals zero obviously just to finish this off we just need to move that 25 over to the other side and i'll get it into the equation of a circle the format that we've looked at throughout the video so if we plus 25 to the other side just to finish this off and then we can get the center and the radius from that so we've got x minus 3 in bracket squared plus y plus 1 in bracket squared and that equals 25. now from there we can get the coordinates of the center and we can also get the radius now the easiest part there well i think they're both relatively simple once you've got to this point but it's nice and easy just to square root this number at the end to get the radius so we know that the radius is going to equal the square root of 25 as that is r squared at the end equals the square root of 25 and in this case that becomes a whole number that becomes 5. okay obviously the square root of a number is positive and negative but we can't have a negative radius so it's just going to be 5 there for the length of the radius now in terms of actually getting the equation of sorry the coordinates for the center now obviously we know it's minus a and minus b on these parts here so a is going to have to be that three right there so the center there we go we know it's going to be three we've taken away three and on this one here we just need to be a little bit more careful because we've got a plus one there so we must have taken away negative one in order for it to turn into positive one and there we go that's our radius and our center taken from that equation okay so there's our two key parts for part a we've got our radius which is five and our center which is three and minus one just need to be careful there when there is a positive number in the bracket like the y plus one that you remember therefore it must have been negative one that we took away in order for it to become positive now the second part here says find the coordinates of the points where c crosses the y axis now in terms of actually thinking about this logically in terms of a coordinate graph if we had a circle let's think about what this would look like so i think the center is at three minus one which is somewhere there and if i was to draw a circle in which i always don't really like doing on the screen here because it's not the easiest to draw but it would look something and it's a really rubbish circle but there we go it looks something like that if i was to draw a basic sketch let's get rid of that center point because that really does not make it look good there we go let's imagine the center there we go three minus one not not to scale but there we go we have these two coordinate points where it crosses through the axes so here and here now the logical thing about this question is that any point on the y axis there the x coordinate down here is always going to be zero so effectively if we want to know where what the y coordinates are where it crosses through the y axis all we have to do is sub in x equals zero let's get rid of that sketch there because it's a little bit of an embarrassing sketch there we go but just to give you a sort of an idea as to what we're actually looking at we're looking at when x equals zero so if i sub that into my equation my equations there i just want to know what does it equal when x equals zero let's put that in so when x equals zero what do we get and i'm subbing it into this equation just here so we get zero minus three in the first bracket so we get negative three squared plus we don't know the y coordinates so we've got y plus one squared and that equals so we just need to rearrange this now negative 3 squared is 9. so actually i could just think that that's 9 there if i take away 9 from both sides we get y plus 1 in bracket squared equals 16 when you take away that 9. now if i square root both sides obviously we want to get y find out what y is so square root both sides we get y plus 1 equals this plus or minus the square root of 16 so it could be plus and minus 4. and obviously then we just need to take away one from both sides and we get y equals minus one plus and minus four and that there will give us two values okay because you could have negative one add four so y could equal 3 or we could have negative 1 take away 4 which would equal negative 5. so we've got 3 or minus 5 in terms of where it crosses the y axis they are the two coordinates there so obviously you could draw a bit more of an accurate sketch now now i know where it crosses through the y axis but there we go that's how we're going to go about solving these two parts of the question so for this question looking at intersecting circles with lines it says show that the line x minus y minus 10 equals zero does not intersect with the circle and then we have our circle equation so we've got our two pieces and obviously the process to solve this will be using simultaneous equations so although we looked at similar quadratic simultaneous equations before this question is slightly different as it says this fact here does not intersect and obviously in this relation we are actually looking at the actual circle so the process to do this is substituting that first equation in so we want to make that x or y equals now look at your circle equation because there are two x's in that equation but there's only this one y squared piece so it would be easier for us if we could make it say y equals so if we take our first equation and add y to the other side we get x minus 10 which is equal to y of course you could write that as y equals x minus 10. so if we take that and substitute it into our circle equation just put in a bracket around x minus 10 when we substitute that in we get x squared minus 4x plus and then we have our x minus 10 which is going to be squared and that is equal to 21. so now we have a an equation that is going to be a quadratic that we're going to need to make equal to zero and then to show that it doesn't intersect we're going to have to show that it has no real roots so having a look at this then let's first expand that double bracket and obviously if you are quick at expanding your double brackets you can obviously just write the answer to that which is x squared minus 20x plus 100 so if we write that all out and simplify it we've got 2x squared so 2x squared we've got a minus 4x minus 20x which is going to be minus 24x we have a plus 100 at the end but we're also going to have to minus this 21 to make sure it equals zero so we can simplify that straight away which is 100 take away 21 which is going to be plus 79 and that is now equal to zero now obviously if they did intersect at this point we would want to solve this equation we'd want to find our x-coordinates and then we would substitute them back in to find our y-coordinates but this says does not intersect so we will look at the discriminant so for the discriminant we are going to look at b squared minus 4ac and then identify our a b and c so a is 2 b is negative 24 and c is 79. and again it's always worth writing them down a is equal to 2 b is equal to negative 24. and c is equal to 79 okay so putting that into b squared minus 4ac we are going to have negative 24 squared and take away 4 times 2 times 79 and if we type that into our calculator and let's just type that in we get negative 24 squared and then minus 4 times 2 times 79 and the answer comes out as minus 56 so we know that when b squared minus 4ac is less than 0 that less than kind of looks a little bit like my c there but that means it has no real roots okay so because we've got an answer of negative 56 we have no real roots and therefore these do not intersect okay so when looking at tangents to circles we just need to remember that the tangent is going to be perpendicular to the radius so for this question here where it says the point p which is 1 minus 2 that lies on the circle with the center 4 6 and it wants us to find the equation of the tangent to the circle at that point we can start by finding the gradient of the radius now what you can do on a question like this is to draw a little diagram which might help you to visualize it and that's fine to do as well it says the center of the circle is 4 6. if we imagine that that is just there and there is a point which is across 1 minus 2 let's imagine that's here and obviously you could draw a sketch of the circle again it doesn't have to be perfect but it might help you to visualize it so we are going to find the gradient of this radius that i've just drawn in and that is between the point four six and the other coordinate down there which is one minus two so obviously you don't need the actual drawing to do that because we can just use our y2 minus y1 over x2 minus x1 so y2 would be 6 take away minus 2 over 4 take away 1. so 6 take away minus 2 is equal to 8 and 4 take away 1 is equal to 3. so the gradient of the tangent is 8 over 3. that means that the gradient of the sorry the gradient of the radius is eight over three now we're going to find the gradient of the tangent because the tangent there is going to have a perpendicular gradient so we are going to find the perpendicular gradient which is the negative reciprocal which is going to be negative three over eight now we can take that and we can put it into the equation of the line given that we know it passes through the point one minus two okay so putting that into the equation of a line we will have y minus y one which is y take away minus two which is going to be y plus two so y plus two is equal to negative three over eight x minus x one and x one is one so x minus one we can now expand that so we have y plus two which is going to be equal to negative three over eight x minus one times minus three over eight is going to be minus 3 over 8 but it was times a negative so it'd be plus 3 over 8. now we can take away 2 from both sides or what we could do here just to make that actually a little bit easier for us is we could just multiply everything by 8 to remove those fractions and that's fine for us to do so we would have 8y plus 16 is equal to minus 3x plus 3. that makes it a lot easier for us to rearrange that and as now we can just minus 16 from both sides or we could move everything to one particular side it doesn't really matter but if we bring this up here in fact if we just move everything to the left then we would have 8y plus 3x when we move that 3x over and also minus the 3 over would give us plus 13 and that is equal to 0. of course you could write this as 8y equals or you could make it y equals if you wanted the question might ask you to write it in different ways but it didn't ask us to write it in a particular way so just having it equal to 0 8y plus 3x plus 13 equals 0 is probably the easiest way for us to write that answer okay so when it comes to triangles in circles there's a couple of different methods that you could use now you might be asked to find the center of the circle now this particular question says to show that a b is the diameter of the circle so of course that we could just find the midpoints of the point a and b but if it wasn't the diameter of the circle and we were wanting to find the center then you could actually use the the perpendicular bisectors of the chords so if we looked at the perpendicular bisector of the chord ac we would find the midpoint and then find the perpendicular line equation which would go through the circle like this we could then do the same with cb finding the midpoint and again the perpendicular bisector would cross through like this now those perpendicular bisectors do cross at the center of the circle so if you found the equation of those two lines you could set them equal to each other and of course we've already discussed looking at midpoints of lines and perpendicular bisectors so that additional step there wouldn't be too much work now this particular question is slightly different it's actually telling us to show that a b is the diameter of the circle now because it that is the diameter we know via circle theorems that if it is this angle here must be a right angle and therefore we could use pythagoras to show this proof so in order to use pythagoras we would have to find the lengths of these lines so a to c and c to b we could then use pythagoras to show the length that it matches the length of a to b so if we go about doing that you can see from a to c the length there goes from minus eight to minus four now that is a distance change of four so we would do four squared and one to nine is a change of eight so if we did four squared plus eight squared and square rooted our answer that comes out as four root five so the length of that line there is four root five we can do the same from c to b now that goes from minus four to four which is a change of eight and nine to five is a change of four so eight squared plus four squared for that length using pythagoras again would give us 4 root 5. so that length there has a length of 4 root 5 which matches the length of a to c so they are our two lengths and if we do pythagoras with those two to find the length of the diameter that should match the length we get when we use this same process so if we just do the same process using pythagoras between these two points then for a to b you can see it goes from -8 to 4 which is a change of twelve and it goes from one to five which is a change of four and if we do that using pythagoras again we get the length for root ten so that comes out as four root ten and now we can have a look at showing our proof so they are the definite distances if this triangle is right angled then if we were to square both of these sides and square root the answer it would have to match the length of the diameter so if we go about doing that four root five in bracket squared plus another four root five in bracket squared and if we square root all of that and we type that into our calculator we get the answer 4 root 10. so as you can see it has matched the length of the diameter therefore it is definitely a right angle triangle and therefore thinking about circle theorems angles made in a semicircle form that right angle that 90 degree angle that we've shown so it must be a right angle triangle and therefore a to b must be the diameter okay so that is the end of the video looking at everything to do with circles i do hope you found that useful and helpful if you did don't forget to like comment and subscribe and i'll see you for the next chapter [Music] you