what annoys me is that some teachers in my school only tells us the formulas and they don't give us an understanding of the topic. Thank you sir you are awesome
I was struggling with this topic from 1 year and after watching this video, i am so so clear! Can't thank Allah enough for making me watch this video..God bless you sir
One doubt. The equation for tangent you have taken is nothing but the slope form, like m=y2-y1/x2-x1 .....but in other questions I have seen for equation of tangent they give answer using the equation ...y=mx+c, where c first we find by substituting x and y from centre (-5,-1) and m of tangent (-2/3) we get c . Then we substitute m and c in y=mc + c to get answer in terms of x and y. So tell me which is correct
Then that method of getting c would be wrong as the centre (-5,-1) does not lie on the line and will therefore not satisfy the equation. The alternative method is to sub the gradient m=-2/3 and the point on the line you can use is (-3,2) as that lies on the line. Sub those values in and c=0
Is there any question about "giving there are tangent to a point on circle,while the question gave u the coordinate of the center and one point from the line (which is NOT the tangent) and the coordinates of the center,and ask u to find out the equation of the tangent line?"
when he tried to find the center and the radius he did it well but i know an easier way. in the end the equation of the circle is the form x squared + y squared +Ax+By+c=0 the center will be (-0.5A,-0.5B) and the radius is 0.5* sq root of A squared + B squared- 4C
@ExamSolutions i see, so first we get the normal gradient and then perpendicular to get the tangent at circles. And In C1 when we do dy/dx of any equation to get the line, is the gradient of the tangent always what we get first? and only then perpendicular for the normal line?
hi sir ! what is the point of contact of tangent to a circle in slope form . like for parabola its a/m^2, 2a/m???? if you answer this it would be really helpfull... !
I understand what you're saying, however what I have is a point on a line which is a tanget to a circle. I have the center point C(4,5), (radius 5) and I have the point on the line P(8,17). The question is; "The line touches the circle at point 'T' and is a tangent. Find the length of PT" I have no idea how to do this... the answers look like they find the distance between the C and T and then use pythagoras therom and subtraction??? Why and how?
Aren't you supposed to use the equation y = mx + c normally...and use coordinates to find 'c', the y intercept. then it would be for example y = 2x + 3 ....idk that's just how I was taught finding equations of lines.
8:03 - You times the "x side" (the right side) by 3.. so why did the numbers in the brackets didn't get multiplied by 3? Does the numbers in the brackets remain as they are?
ah. I understand everything. Just one tiny thing im extremely confused about. Is when getting the gradient of something, i always always mix up and am confused about which one is the tangent gradent, which one is the normal gradient.
I'm having trouble doing a question where I'm given 2 circles and asked to find the common tangents. Is there a video here explaining how to do that? Any help would be appreciated.
Hey I have a question, I could easily do this with implicit differentiation, taking me less than 20 second. However thats c3-4 knowledge, if i do it in the exam will they mark me down or will i be awarded in full?
@@dsrmanav5560 I asked that question 7 years ago I've just graduated today with a Masters in Mechanical engineering. I think now I'm ready to answer some of the questions you might have 😂
M Shezz to make the equation into the form of a circle equation he added 25 and 1 to both sides and took away 13 from both sides with overall means 13 is left on the right side as r^2
when finding the equation of CP you used the formula y2-y1/x2-x1... how do you know which coordinates are y2 and x2, and which are the y1 and x1. I got a gradient of 3/2 for CP so the gradient of the tangent became -2/3 and you got 2/3. I thought x1 and y1 were the coordinates at point P but you used them as y2 and x2????
hi im a little confused, i did a C2 paper the other day(June 2010 Q.10 (b)) and it said L1 is a tangent to circle C at B. find equation... in the answer the poduct of the gradient had to multiply to give -1. however in a C3 june 2008 Q.1 (b) to find the gradient to the tangent the product of the gradients never had to equal -1 i.e i only differentiated to get the gradient and no further
They should teach the y-y1=m(x-x1) formula to GCSE students more. In school they only taught me y=mx+c and now I have to learn the y-y1 formula from scratch as its so much better and more useful for A level maths.
I know a lot of people probably told you this but you are the best math teacher a student could ask for....period.
Thanks a Lot
what annoys me is that some teachers in my school only tells us the formulas and they don't give us an understanding of the topic. Thank you sir you are awesome
thanks you awesome
You can also use the equation y=mx+c I find it much easier to use
You. Are. THE BEST!
your videos are very helpful simple and they straight forward
I was struggling with this topic from 1 year and after watching this video, i am so so clear! Can't thank Allah enough for making me watch this video..God bless you sir
Awesome was stuck on this question for a while, great teaching aswell 👍
thank you so much , my maths hero !!!!!
Thanks
Loved the other video, but do u have anyone on normal to a circle
I appreciate your help! You saved me; I am now able to get a pass on my exam!
Very well explained!
That brush 🖌️❤️
Thank you sir... the way you teach through this makes it easier to comprehend because you teach slowly and step-by-step.. in exam style.
If you differentiate a curve it gives you the general equation of a tangent not a normal or perpendicular.
One doubt. The equation for tangent you have taken is nothing but the slope form, like m=y2-y1/x2-x1 .....but in other questions I have seen for equation of tangent they give answer using the equation ...y=mx+c, where c first we find by substituting x and y from centre (-5,-1) and m of tangent (-2/3) we get c . Then we substitute m and c in y=mc + c to get answer in terms of x and y. So tell me which is correct
Then that method of getting c would be wrong as the centre (-5,-1) does not lie on the line and will therefore not satisfy the equation. The alternative method is to sub the gradient m=-2/3 and the point on the line you can use is (-3,2) as that lies on the line. Sub those values in and c=0
wow, thanks. Forgot the circle equation a long time ago
Is there any question about "giving there are tangent to a point on circle,while the question gave u the coordinate of the center and one point from the line (which is NOT the tangent) and the coordinates of the center,and ask u to find out the equation of the tangent line?"
when he tried to find the center and the radius he did it well but i know an easier way. in the end the equation of the circle is the form x squared + y squared +Ax+By+c=0 the center will be (-0.5A,-0.5B) and the radius is 0.5* sq root of A squared + B squared- 4C
Perfectly done.
@cardboardlife Its one term on the right so the 3 gets cancelled.
@ExamSolutions i see, so first we get the normal gradient and then perpendicular to get the tangent at circles.
And In C1 when we do dy/dx of any equation to get the line, is the gradient of the tangent always what we get first? and only then perpendicular for the normal line?
@rustytoe178 For the x-axis set y=0, for the y-axis set x=0
Thank you. I have worked out the equation of the tangent. How do i then calculate the points where the tangent crosses the axes? Thanks.
hi sir ! what is the point of contact of tangent to a circle in slope form . like for parabola its a/m^2, 2a/m???? if you answer this it would be really helpfull...
!
Great Video! But would've liked to see how to get the equation of a tangent when given the circles equation, and the gradient of a parralel tangent.
@syndrome21 No probs - pleased to hear that it helped
@bollywoodmassive not necessarily, only if they ask.
thanks you a lot for a good explaination.
Very kind, thank you.
@pinxify05 That's ok. Not sure though what plates are though.
I understand what you're saying, however what I have is a point on a line which is a tanget to a circle. I have the center point C(4,5), (radius 5) and I have the point on the line P(8,17). The question is; "The line touches the circle at point 'T' and is a tangent. Find the length of PT" I have no idea how to do this... the answers look like they find the distance between the C and T and then use pythagoras therom and subtraction??? Why and how?
You are right but I think the center should be ( -5,1)
@MrinalChoudhary That's good.
very helpful. thanks a million!
To find the centre I just use (-g,-f), with g being the number in front of the x and f being the number in front of the y.
g would be the coefficient of x divided by 2 and f would be the coefficient of y divided by 2
Aren't you supposed to use the equation y = mx + c normally...and use coordinates to find 'c', the y intercept. then it would be for example y = 2x + 3 ....idk that's just how I was taught finding equations of lines.
it also helps students like me doin additional maths, thanx a lot :D :D :D
Not really. You can use either but y-y(1)=m(x-x(1)) is more efficient.
Is it possible to differentiate the equation you were given and then put in -3 to find the gradient of the tangent?
Yes
@lpspgameplay The normal gradient goes from the centre to the circumference and the tangent is the perpendicular to this.
superb - thanks
appreciated, thanks
8:03 - You times the "x side" (the right side) by 3.. so why did the numbers in the brackets didn't get multiplied by 3?
Does the numbers in the brackets remain as they are?
ah.
I understand everything. Just one tiny thing im extremely confused about.
Is when getting the gradient of something, i always always mix up and am confused about which one is the tangent gradent, which one is the normal gradient.
@ExamSolutions Thank you!
Thanks man 💓💓
No problem.
I had pre-worked it out.
Thanks for ur video...its really helpful for me to understand it easily..^^
thanks a lot to this. it is such a great help doing my plates on analytic geometry :]
I'm having trouble doing a question where I'm given 2 circles and asked to find the common tangents. Is there a video here explaining how to do that? Any help would be appreciated.
Nikotine97 Sorry know. Put the question up on facebook.com/groups/566605273474018/ and you will get it answered.
what is the purpose of a tangent line
Could do do a vid 4 ( the equation of a tangent from a point outside a circle) please?
Hey I have a question, I could easily do this with implicit differentiation, taking me less than 20 second. However thats c3-4 knowledge, if i do it in the exam will they mark me down or will i be awarded in full?
You should be awarded this in full. I would check this out with your teacher as well.
Alright, thanks alot!
hi
can i differentiate implicitly and substitute the values (-3,2) to get the gradient?
@@dsrmanav5560 I asked that question 7 years ago I've just graduated today with a Masters in Mechanical engineering. I think now I'm ready to answer some of the questions you might have 😂
@@dsrmanav5560 😂😂😂😂😂😂 keep it up bro let me know if you need help with maths tho 😭😭😭😭
@@dsrmanav5560 or physics 😭😭😭
@@dsrmanav5560 ❤️❤️❤️
by Differentiation we can find tangent in short steps but thanks alot!!!
what if the center of a circle is given and equation of a tangent to a circle is given.......and we should find the equation of the circle
what if the points (3,2) is not given and it says 'it cuts the x-axis/y-axis' ? what should i do then? plz help thnx :)
@TheSwiftmeister What do you want to know?
Thank you !!!
No problem - pleased to hear it helped.
Shouldn't the equation be in terms of y=mx+c
+Tejdeep Shinh No, not necessarily. You can have it in any form unless asked to put it in the form y=mx+c or ax+by+c=0.
+Tejdeep Shinh you can use both
thanx a lot..really helpful..:)
can i ask why the numberrs outside the brackets are negative and not positive because when u square them how would u get a negative number
Sorry but I am lost on your question.
ExamSolutions no is ok thank u I rewatched the video n it was answered thank u the video was verrry useful
@@richardwarlocc238 No problem. Good to hear it is sorted.
Shouldn’t that 13 of been a -13 when you move it over to the other side of the equals the sign has to change. No?
M Shezz to make the equation into the form of a circle equation he added 25 and 1 to both sides and took away 13 from both sides with overall means 13 is left on the right side as r^2
This helped so much! Thank you :D
when finding the equation of CP you used the formula y2-y1/x2-x1... how do you know which coordinates are y2 and x2, and which are the y1 and x1. I got a gradient of 3/2 for CP so the gradient of the tangent became -2/3 and you got 2/3. I thought x1 and y1 were the coordinates at point P but you used them as y2 and x2????
more analytic geometry vids pls. Thanks :D
hi im a little confused, i did a C2 paper the other day(June 2010 Q.10 (b)) and it said L1 is a tangent to circle C at B. find equation... in the answer the poduct of the gradient had to multiply to give -1. however in a C3 june 2008 Q.1 (b) to find the gradient to the tangent the product of the gradients never had to equal -1 i.e i only differentiated to get the gradient and no further
Thank you!!!!!!!!!!!!!!!!!!!!
thanks m8
@wenger26 Cheers
this is so helpful thank you. :)
thanks a lot bro
That's okay.
@lpspgameplay Yes
@werty123x Thanks for the complement
Nice one:)
what will i get when I times -1\4 to x-2? help please thank you
+Angela Javier -x/4 +1/2
@SulthanaB226 Keep working hard and hopefully you will get it.
Arigato gozaimashita.
Nice vid
@grimestarr Thanks and Good Luck when it comes.
is this A-Level? cuz its similar to gcse stuff...
It was designed for A-level.
They should teach the y-y1=m(x-x1) formula to GCSE students more. In school they only taught me y=mx+c and now I have to learn the y-y1 formula from scratch as its so much better and more useful for A level maths.
how do you know that the circle is in that exact position on the diagram and the tangent is negative not positive?
Didn’t know Gary Lineker was a maths teacher...
@ktbMonster Cool
why is gradient -2/3 when it says tangent not normal?
The tangent for a circle is perpendicular to the gradient of the radius CP for a circle.
ExamSolutions thanks
cool
Wtf... I have to do this for GCSEs.......
new spec gl LUL
C1 and some of C2 have very basic math, this happens to be one of them. The more complex stuff come when you do C3 and C4
That's peak.
@Brianyoutubation A good idea. Care to make a donation and send me an email.
Sir, am I the only who thinks C2 is easier than C1?
Bro why didnt your +13 change to -13 when u switched it to the other side
And why is there a minus 1 in the center buh plus 1 in the working
On the left hand side it was -25-1+13 which is -13 so then I added 13 to both sides.
Wau
this is useful and all, but real exam questions compared to this is just...
I will in my boots!
Seems I'm a bit late
Yup. I'm dumb