ITS MY HIGHER MATH EXAM TOMORROW! HOW THE HELL DO I DO THISs?

Question

The point P(2,3) lies on the circle (x+1)^2 + (y-1)^2 = 13

Find the equation of the tangent at P.


Whoever shows me how to do this the best gets a best answer and all my looove.


Thank you soooo much x x x x

Answer 1

I think probably the easiest way is to remember that a radius of a circle to a point is perpendicular to the tangent line at that point.

You know the center of the circle, so you could find the slope of the radius from the center (-1, 1) to the point (2, 3).

Use the fact that perpendicular lines have negative reciprocal slopes, and you will know the slope of your tangent line at that point.

Use the slope and the point to find the equation of the line.

Hope that helps!

Answer 2

Centre of the circle is (-1 , 1)

Gradient of the radius from the centre to (2 , 3) is

m = (3 - 1) / (2 - -1)

m = 2/3

gradient of tangent will be - 3/2 as it is perpendicular to the radius

equation of the tangent is y = (-3/2)x + c but it goes through (2,3)

so 3 = 2.(-3/2) + c .....so c = 6

equation is y = (-3/2)x + 6 or 2y = -3x + 12

Answer 3

Solve the equation for y
(y-1)^2 = 13 - (x+1)^2
y = 1 ± √[ 13 - (x+1)^2 ]

Take the derivative:
dy/dx = ±(1/2)[ 13 - (x+1)^2 ]^(-1/2) * [-2(x+1)]
dy/dx = ±[ 13 - (x+1)^2 ]^(-1/2) * [(x+1)]

The point (2,3) lies in the top right of the circle, so the slope is going to be negative. So take the negative value here when x=2.

dy/dx = -[ 13 - (2+1)^2 ]^(-1/2) * [(2+1)]
dy/dx = -3[ 13 - (3)^2 ]^(-1/2)
dy/dx = -3[ 4 ]^(-1/2)
dy/dx = -3/2

Answer 4

Hope you have time to look at this quick solution:-
Centre A(-1,1)
m AP = (3 - 1) / (2 + 1) = 2 / 3
m tangent = - 3/2
Tangent passes thro` (2,3)
y - 3 = (- 3/2).(x - 2)
y = (- 3/2).x + 6

Good luck to- day.

Answer 5

Evaluate the first derivative at that point.
Chain rule.
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