Mathematics, 17.01.2020 07:31 gafsh272833

Sketch the circle of radius 2 centered at (3,–3) and the line l with equation y = 2x + 2 find the coordinates of all the points on the circle where the tangent line is perpendicular to l.

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Answer from: rleiphart1

The points on the circle where the tangent line is perpendicular to L are,

$(-3-\dfrac{4\sqrt{5}}{5}, -3+\dfrac{4\sqrt{5}}{5})$

The points of the circle (3,-3)

The equation of the line is,

y=2x+2

The standard equation of the cirlce with point (h,K) and radius r can be given as,

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

(x-3)^2+(y+3)^2=2^2

(x-3)^2+(y+3)^2=4

Find p points of the other circle with the following equation,

y-(-3)=2(x-3)

y=2x-9

Put this value of y in the equation of circle,

(x-3)^2+(2x-9+3)^2=4

x^2-6x+9+4x^2-24x+36=4

5x^2-30x+41=0

On solving the above equation we get the two roots of x,

$x=3-\dfrac{2\sqrt{5}}{5}$

$x=3+\dfrac{2\sqrt{5}}{5}$

Put this value of x to find out the value of y.

y=2x-9

$y=2(3-\dfrac{2\sqrt{5}}{5})-9$

$y=-3-\dfrac{4\sqrt{5}}{5}$

Similarly taking negative sign we get,

$y=-3+\dfrac{4\sqrt{5}}{5}$

Hence the points on the circle where the tangent line is perpendicular to L are,

$(-3-\dfrac{4\sqrt{5}}{5}, -3+\dfrac{4\sqrt{5}}{5})$

For more about the circle follow the link below-

link

Answer from: liddopiink1

$(3-\frac{2\sqrt{5}}{5},-3-\frac{4\sqrt{5}}{5})$ and $(3+\frac{2\sqrt{5}}{5},-3+\frac{4\sqrt{5}}{5})$

Step-by-step explanation:

We are asked to graph the circle of radius 2 centered at (3,–3) and the line L with equation y=2x+2 .

We know that standard form of circle with center at point (h,k) is (x-h)^2+(y-k)^2=r^2 . Upon substituting our given values, we will get:

(x-3)^2+(y-(-3))^2=2^2

(x-3)^2+(y+3)^2=4

Please find the attachment for the graph.

We need to find the points, where, the tangent of circle are perpendicular to L.

We know that slopes of perpendicular lines are negative reciprocal of each other, so slope of tangent line would be negative reciprocal of 2 that is $-\frac{1}{2}$ .

We need to find P points on the circle such that the radial line from point (3,–3) to P has a slope equal to 2.

$\frac{y-(-3)}{x-3}=2$

$\frac{y+3}{x-3}=2$

y+3=2x-6

y=2x-9

Now, we will substitute this equation in circle equation to solve for x-coordinates of points.

(x-3)^2+(2x-9+3)^2=4

(x-3)^2+(2x-6)^2=4

x^2-6x+9+4x^2-24x+36=4

5x^2-30x+45=4

5x^2-30x+41=0

Now, we will use quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-30)\pm\sqrt{(-30)^2-4(5)(41)}}{2(5)}$

$x=\frac{30\pm\sqrt{900-820}}{10}$

$x=\frac{30\pm\sqrt{80}}{10}$

$x=\frac{30\pm4\sqrt{5}}{10}$

$x=\frac{30-4\sqrt{5}}{10}\text{ or} x=\frac{30+4\sqrt{5}}{10}$

$x=3-\frac{2\sqrt{5}}{5}\text{ or} x=3+\frac{2\sqrt{5}}{5}$

We figured our x-coordinates of our required points.

Now, we will substitute these points in equation y=2x-9 to find y-coordinates as:

$y=2(3-\frac{2\sqrt{5}}{5})-9$

$y=6-\frac{4\sqrt{5}}{5})-9$

$y=-3-\frac{4\sqrt{5}}{5})$

2nd y-coordinate:

$y=2(3+\frac{2\sqrt{5}}{5})-9$

$y=6+\frac{4\sqrt{5}}{5})-9$

$y=-3+\frac{4\sqrt{5}}{5}$

Therefore, our required points are $(3-\frac{2\sqrt{5}}{5},-3-\frac{4\sqrt{5}}{5})$ and $(3+\frac{2\sqrt{5}}{5},-3+\frac{4\sqrt{5}}{5})$ .

Answer from: Quest

$r(x)=-\frac{1}{50}x^2+14x$

step-by-step explanation:

let's assume

number of candles set =x

price of of x candles =p(x)

we are given

a candle maker prices one set of scented candles at $10 and sells an average of 200 sets each week

so, we get point as (200,10)

$x=200,p=10$

he finds that when he reduces the price by $1, he then sells 50 more candle sets each week

so, we can find slope

$m=-\frac{1}{50}$

now, we can use point slope form of line

$p(x)=mx+b$

now, we can plug values

$p(x)=-\frac{1}{50}x+b$

we can use point

$x=200,p=10$

and find b

$10=-\frac{1}{50}\times 200+b$

$10=-4+b$

$b=14$

now, we can plug it back

and we get

$p(x)=-\frac{1}{50}x+14$

now, we can find revenue

we know that

revenue = price*quantity

so, we get

$r(x)=x\times p(x)$

$r(x)=x\times (-\frac{1}{50}x+14)$

$r(x)=-\frac{1}{50}x^2+14x$

Answer from: Quest

let f (x ) is a function and c be a positive real number, the graph of the function g (x ) = f (cx ) is represented as follows.

when, c > 1, the function compresses it in the x - direction.

when, 0 < c < 1, the function stretches it.

the function is g ( x ) = (9x )2.

the heighest power of x in g ( x ) = (9x )2 is 2, so the parent function is square function.

the equation of parent function is f ( x ) = x 2 .

the function is g ( x ) = (9x )2 and the parent function is f ( x ) = x 2 .

so, the functions f and g have the following relationship.

g ( x ) = f ( 9x ).

compare the function g ( x ) = f ( 9x ) with g (x ) = f (cx ).

c = 9 ( > 1 ).

so, the function g (x ) = (9x )2 compresses it in the x - direction.

step-by-step explanation:

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Sketch the circle of radius 2 centered at (3,–3) and the line l with equation y = 2x + 2 find the co...

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