If x^2 + y^2 - 2x - 4y = 20 , then an equation of the tangent line at (5, -1) could be one of the following..?

Question

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3x + 4y - 36 = 0

4x + 3y - 35 = 0

3x - 4y + 30 = 0

4x - 3y - 23 = 0

3x + 4y + 14 = 0

4x + 3y + 15 = 0

3x - 4y - 20 = 0

4x - 3y + 27 = 0

Answer 1

Implicit differentiation:
2x + 2yy' - 2 - 4y' = 0
Solve for y':
(2y-4)y' = 2-2x, so y' = (2-2x)/(2y-4) = (1-x)/(y-2).

Plug in x=5, y=-1, and get y' = -4/3.

Therefore, the equation of the tangent line, using the point-slope form, is

y - (-1) = (-4/3)(x-5), or

y+1 = -4/3x + 20/3
Multiply by 3:
3y + 3 = -4x + 20
Rearrange:
4x + 3y - 17 = 0
so it's none of the above.

Answer 2

Firstiful this is an equation of a circle with a centre of (1,2) and has a radius of 5. Therefore the orginal eqn is (x-1)^2+(y-2)^2=25.
Then rearrange the eqn above to y = (-x^2+2x+24)^1/2+2. Find the eqn of the grad using the one we just did, so we got
y'= (1-x)(-x^2+2x+24)^-1/2 then sub x = 5 into the eqn to find out the grad at the pt of (5,-1). We then find out the gradient is + or - 4/3, however we going to take the + one as the tang at that pt is going up (draw a pic and see).

Now use a Pt-grad formula y-y1=m(x-x1), sub m=4/3 with a pt (5,-1) into the formula, you then should get 4x-3y-23=0 (gen form)

Answer 3

James' method will work well but you need to check the value for y' at (5,-1) - good luck - Mike

Answer 4

the answer is 56