I see you have two separate questions/problems here. The first ones is about the circle with parts a and b. The second one is about the derivative of f(x).Originally Posted bydadon

Here are some ways to solve them.

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A) Circle 2x^2 +2y^2 -4x +10y = 8 -----------(i)

a)find the center and radius.

To find the center and radius, we need to convert the general form, (i), into its standard form

(x-h)^2 +(y-k)^2 = r^2 -------------(ii)

where

(h,k) is the center

r is the radius.

We do that by applying the "complete the square" on (i):

2x^2 +2y^2 -4x +10y = 8 --------(i)

[2x^2 -4x] +[2y^2 +10y] = 8

We do completing squares only when the squared variable has 1 as its coefficient, so,

2[x^2 -2x] +2[y^2 +5y] = 8

[x^2 -2x] +[y^2 +5y] = 4

[x^2 -2x +(2/2)^2 -(2/2)^2] +[y^2 +5y +(5/2)^2 -(5/2)^2] = 4

[(x-1)^2 -1] +[(y +2.5)^2 -6.25] = 4

(x-1)^2 +(y+2.5)^2 = 11.25 ----------(iia)----in the form of (ii).

Therefore, center is (1,-2.5), and radius is sqrt(11.25) = 3.3541 units. ----answer.

b) I assume that you mean the tangent "where", not "when" as posted, the circle intersects the x-axis in the first quadrant.

Because it it is "when" then there infinitely many possible tangent lines.

Where the circle intersects the x-axis, the y is 0, so, using (iia),

(x-1)^2 +(0+2.5)^2 = 11.25

(x-1)^2 +6.25 = 11.25

(x-1)^2 = 5

x-1 = +,-sqrt(5)

x = 1 +sqrt(5) ----------in the 1st quadrant.

So the point in question is (1+sqrt(5),0). ----------***

The slope of the tangent line at any point is dy/dx or y', so differentiating both sides of (iia) with respect to x,

(x-1)^2 +(y+2.5)^2 = 11.25 ----------(iia)

2(x-1) +2(y+2.5)*y' = 0

y' = -(x-1)/(y+2.5)

Plugging the coordinates of the point in question,

y' = -(1+sqrt(5) -1) / (0 +2.5)

y' = -sqrt(5) /2.5 -------------the slope of the tangent line.

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

y -0 = -sqrt(5)[x -(1+sqrt5)]

y = -sqrt(5)[x -1 -sqrt(5)] ------------answer.

Or, in its y = mx +b form,

y = -sqrt(5)*x +(5+sqrt(5)) -----------answer.

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B) The derivative of

f(x) = Square root((x+1)/(x-1)) ------(1)

To avoid the quotient rule, if you are having trouble with it, is to use the product rule.

Eq.(1) can be expressed as

f(x) = [(x+1)*(x-1)^(-1)]^(1/2)

f(x) = (x+1)^(1/2) *(x-1)^(-1/2)

Then,

f'(x) = [(x+1)^(1/2)]*{(-1/2)*(x-1)^(-3/2)} +[(x-1)^(-1/2)]*{(1/2)*(x+1)^(-1/2)}

Simplifying that, you should get

f'(x) = -(x+1)^(-1/2) *(x-1)^(-3/2) ---------answer.