Equation of tangent to Circle x^2+y^2-8x-10x-128=0 IB Math GCSE
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- Опубликовано: 17 сен 2024
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This is very clear explained...i really love it ....because it will help me with the exam coming tomorrow.
To find center of a circle in standard form and general form like the one used in video, just take derivative w.r.t x or y term of circle.
For example: x²+y²-8x-10y-128=0
For center C(h,k), i need h so i will take derivative w.r.t x. So,
2x+0-8-0-0=0. (EVERYTHING except x will be treated as constant term) (remember it is for this scenario only that we treat y as constant and not y' )
2x=8. => x=4
For k, take derivative w.r.t y,
0+2y-0-10-0=0
2y=10. => y=5
Hope it helps.
that is sick
More easy method
Just multiply the coefficient of x and y by -1/2 , to get coordinates of cente
You don't need to differentiate at all!
Thanks but how then do you find the radius?
Thanks..... you're the best..... you've explained excellently
Call me overly cautious, but I would first check that the desired point of tangency is actually on the circle (I checked. It is).
Correct
Thanks so much Mr.Kumar
Thank you very much sir I must say I enjoyed your lecturing, ❤❤❤ your English is amazing
Thanks for appreciation
You're very smart at explanations thank you've really helped me 🙏🙏🙏
thank you so much this man is saving lives
This really helped me, thank you
Very helpful 🙏🙏🙏🙏🙏
cleared all the doubts Thanks a lot Anil sir
Thank so much for your help
thank you prof God Bless
best ever
Thankyou sir you explained very well🙏🏻
very helpful sir!!
Nice explanation
Thank you Sir that was really helpful.
Thanks for appreciation
Hello, I'm having a question which is more of the same like that but a bit different,. Find an equation of each tangent line drawn to from the point (5,1) to the circle x^2 + y^2 = 14. Please I need my feedback.
Yep, this one please!!
5x+y=14
So useful!
Thanks!😁👌
Very helpful 💚
Thanks it really helped
dude you saved me, questions can we send fanmail?
Sir, shall I use to get tangent equation directly from given circle equation by this way: x*{-8} + y*10 - 4{x-8} -5{y+10} -128=0 which implies 12x - 5y +146 = 0
So does it mean that the equation of the normal line to the circle is the same as the equation of the radius
Thanx sir
Nice one
Thanks
Please sir how did you know that the tangent line will be drawn at that point
Thank you sir
Thanks so much,
Hello sir .... please share few qs related circle .. which possibility comes in exams
Find the equation of the line passing through the center of the circle x^2+y^2-4x-2y-6=0 and the point (3, 2)
2x + 3y + 4)(2x + 3y - 5) = (ax2 + by2 + 2hxy + 2gx + 2fy + c) how we can find the value of a b h g f and c faster sir if you know then tell me in comment i get the answer sir please clear my doubt
-5² is + 25 right? The radius mas be the squareroot of 119
Can you write the final equation starting with 'y' instead of '0' ?
Yes, you can. Thanks
Sir it's y-5=12/5 (x-4)
Last step
Thank you, but i don't understand well
We found the centre by completing the squares first. Then the equation of tangent as it is perpendicular to the radius. Hope that helps. Thanks
M1 =12/5 how
Obtain the equation of the tangent to the parabola
y 4x
2
which is parallel to the line
y 2x 0
and find the coordinates of the point of contact