# I've a maths question that I cannot solve

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#1
"L is the circle with equation x^2 + y^2 = 4.
P{3/2,√7/2} is a point on L.

Find an equation of the tangent to L at the point P."

I would very much appreciate a step-by-step guide where every step is explained on how to answer this question, and thanks in advance to anyone who manages to solve it and present their working and reasoning.
1
3 years ago
#2
solve to find y so y=(4-x^2)^1/2 (square root) then differentiate for gradient so m=x^3-4x using chain rule then sub in value 3/2 from P and finally sub in all values into y-y1=m(x-x1)
0
3 years ago
#3
IDK if this is the correct method or whatever, but I would find the gradient of the line with the points (0,0) (3/2,√7/2). Then I would find the negative reciprocal of that, as a tangent to a circle meets the line of the radius at 90 degrees, so they are perpendicular. Put whatever the negative reciprocal is into the equation for a line, y = mx + c. Then substitute the points of P into it, and re-arrange to find C.
1
3 years ago
#4
(Original post by mathsnerd49)
solve to find y so y=(4-x^2)^1/2 (square root) then differentiate for gradient so m=x^3-4x using chain rule then sub in value 3/2 from P and finally sub in all values into y-y1=m(x-x1)
Isn't differentiating (or whatever it is) A-Level?
0
3 years ago
#5
(Original post by Bill Nye)
Isn't differentiating (or whatever it is) A-Level?
I did it in year 11 but it may be now, idk what method you'd want without differentiation
0
#6
(Original post by mathsnerd49)
solve to find y so y=(4-x^2)^1/2 (square root) then differentiate for gradient so m=x^3-4x using chain rule then sub in value 3/2 from P and finally sub in all values into y-y1=m(x-x1)
Wow! I've never seen such maths, and I don't think I will for a while, seeing as I'm only moving from year 9 into 10, and though it looks impressive, I have no idea what's going on. While I do thank you greatly for taking the time to answer my question, I feel like I should be clearer with my question. As such, I shall upload the mark scheme. Do you think you could explain to me what it's doing and why?
0
#7
(Original post by Bill Nye)
IDK if this is the correct method or whatever, but I would find the gradient of the line with the points (0,0) (3/2,√7/2). Then I would find the negative reciprocal of that, as a tangent to a circle meets the line of the radius at 90 degrees, so they are perpendicular. Put whatever the negative reciprocal is into the equation for a line, y = mx + c. Then substitute the points of P into it, and re-arrange to find C.
I think you just explained the mark scheme. Thank you very much!
0
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