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3 years ago

Find equation tangent to a circle at given point

1 answer:

7 0

the equation of the tangent line must be passed on a point A (a,b) and perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle, its center is C(0,0). And we assume that the tangent line passes to the point A(2.3).

</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.

<span>Let's find the equation of the line parallel to the radius.</span>

<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>

det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.

let's find the equation of the line perpendicular to this previous line.

let M a point which lies on the line. so MA.AC=0 (scalar product),

it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent

You might be interested in

15.0 kg mass is displaced 3.00 m south and then 4.00 m west by a 10.0 N force. What is the total work done on the object?

masha68 [24]

<h2>Answer: D. 70.0 J</h2>

Explanation:

3.00 m + 4.00 m = 7.00 m x 10.0 N = 70.0 J

5 0

3 years ago

Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days

Vinvika [58]

your question is missing the given conditions, here is the complete question;

Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average time A and B can complete the task working alone is 12.5 days.

(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task

Answer:

Machine A, working alone, can complete the task in 15 days.

Explanation:

As the average is given as 12.5 so,

A+B=2*12.5

A+ B=25

second condition says machine A takes 5 more days to complete the task alone than B so

A=B+5........Eq1

Now from the question statement we can get that

1/A+1/B=1/6, where A is time needed for A to do the task alone and B is the time needed for B to complete the task alone

thus, simplyfing above equation we get as

AB/A+B=6; \\ putting A+B=25 we get

AB/25=6

AB=6*25

AB=150; \\ putting Eq1 we get

(B+5)B=150

B^2+5B-150=0

simplifying the quadratic equation above we get B=10

putting B=10 in Eq1, we get

A+10=25

A=15

Thus, Machine A, working alone, can complete the task in 15 days.

3 0

3 years ago

Problem 3) Bob stands at the edge of the swimming pool holding a laser 1.5m above the ground. He shines the red laser beam onto

Vadim26 [7]

Answer:

d = 5.75m

Explanation:

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

Here,

i = 90 - θ

$\theta = \tan^-^1(\frac{1.5}{3} )\\\\=26.56^\circ$

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

$r = \sin^{-1}\frac{(1)\sin (90-26.56)}{1.33}\\\\r = 42.26m$

$\tan r = \frac{2.5}{d_1}$

$d_1 = \frac{2.5}{\tan (42.26)} \\\\d_1 = 2.75m$

Therefore, the distance is

d = 3 + d₁

d = 3 + 2.75

d = 5.75m

6 0

3 years ago

Two blocks are connected as shown in the diagram below. Assume that the ramp is frictionless. Draw the force diagram for the blo

cluponka [151]

Answer:

diagram: see image, x-component: 84.3 N, acceleration: 4.38 m/s^2

Explanation:

(see image for further explanation)

5 0

3 years ago

A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit

Tom [10]

1.97 E-7 m should be right. I took this a while ago.

6 0

3 years ago