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

v = 1/3 a^2h

step-by-step explanation:

the volume is 1/3 a^2h.

Answer from: Quest

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.

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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...