Line l is tangent to circle O at point P(3,4) where the center is located at (0,0).

a. Find the radius of the circle. (I got 4)
b. Give an equation of the circle. (I got x^2+y^2=16)
c. Find the slope of line l.
d. Give an equation of line l.
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I can't figure out what the slope would be, if I knew that I could figure out (d). Does anybody know how to do this?

Since the line is tangent at a point to the circle, then the point lies on the circle. Also, since you know the center, the radius is simply the distance between the points (0,0) and (3,4).

a) So, radius=d=sqrt((3-0)^2+(4-0)^2)=5
b) x^2+y^2=25
c) Observe that the radius is perpendicular to the tangent line. Thus, if you can find the slope using the coordinates of endpoints of the radius (i.e. (0,0) and (3,4)) then the negative reciprocal of this value will give the slope of the tangent line. Answer is -3/4.
d) Since you know the slope (-3/4) and the point (3,4) is on the line, we can use the form y=mx+b. If you go through the motions, you'll find that y=(-3/4)x+(25/4)

Given the points (-3,4) and (0,0) find the equation of the radius, the equation of the tangential line and equation of the circle.

Radius: The radius will be the line connecting the center of the circle to the point where the tangent touches the circle. Since the center of the circle is (0,0) and the point of tangency is (0,0), we just need to find the distance between these two points.

Distance between (-3,4) and (0,0):
d = √[(0-(-3))^2 + (0-4)^2]
d = √[9 + 16]
d = √25
d = 5

So the radius has length 5, and its endpoints are (0,0) and (-3,4). The equation of the radius is:

y = (4/(-3))x + 0
y = (-4/3)x

Tangential Line: The tangent line will pass through the point of tangency, which is (0,0). To find the slope of the line, we need to find the perpendicular slope to the radius. Since the radius has slope -4/3, the slope of the tangent line will be the negative reciprocal of -4/3:

m = 3/4

So the equation of the tangential line will be:

y = (3/4)x + 0
y = (3/4)x

Circle: We can use the midpoint formula to find the center of the circle, which will be the midpoint between (-3,4) and (0,0).

x_mid = (-3+0)/2 = -1.5
y_mid = (4+0)/2 = 2

So the center of the circle is (-1.5,2), and the radius is 5. Thus, the equation of the circle is:

(x+1.5)^2 + (y-2)^2 = 25

There are four points on the grid, (-3,0+_4) and (-3+_3,0) and C(-2,-4)

A) What is the equation of the line segment A(-6,0) B(-3,4)
B) Find the shortest distance from B to the center of the shape created.
C) calculate the area covered by the shape
D) determine the size of angle ABC where the coordinate of point C is (0,0)

A) The equation of the line segment AB can be found using the two-point form of the equation of a line:

(y - y1)/(x - x1) = (y2 - y1)/(x2 - x1)

Substituting the values of the two points A and B:

(y - 0)/(x - (-6)) = (4 - 0)/(-3 - (-6))

Simplifying this equation:

y/(x + 6) = 4/3

Cross-multiplying:

3y = 4x + 24

So the equation of the line segment AB is:

4x - 3y + 24 = 0

B) To find the center of the shape created by the four points, we can find the midpoint of segment AC and the midpoint of segment BD, and then find the midpoint of those midpoints.

Midpoint of AC:

x_mid1 = (-3 - 2)/2 = -5/2
y_mid1 = (0 - 4)/2 = -2

Midpoint of BD:

x_mid2 = (-3 + 0)/2 = -3/2
y_mid2 = (4 + 0)/2 = 2

Midpoint of midpoints:

x_mid = (-5/2 - 3/2)/2 = -2
y_mid = (-2 + 2)/2 = 0

So the center of the shape is (-2,0).

To find the shortest distance from B to the center, we can use the distance formula:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Substituting the values for B and the center:

d = sqrt((-2 - (-3))^2 + (0 - 4)^2)
d = sqrt(1 + 16)
d = sqrt(17)

So the shortest distance from B to the center is sqrt(17).

C) To find the area of the shape created by the four points, we can divide it into two triangles and find their areas. One triangle is formed by points A, B, and C, and the other triangle is formed by points B, C, and D.

Triangle ABC:

Base = 3 (distance between A and B)
Height = 4 (distance from B to the x-axis)

Area = (1/2) * 3 * 4 = 6

Triangle BCD:

Base = 1 (distance between B and D)
Height = 4 (distance from B to the x-axis)

Area = (1/2) * 1 * 4 = 2

Total area = 6 + 2 = 8

D) To find the size of angle ABC, we can use the dot product formula:

a · b = |a| * |b| * cos(theta)

where a and b are vectors, |a| and |b| are their magnitudes, and theta is the angle between them.

We can use vectors BA and BC. The magnitude of BA is sqrt((3 - (-6))^2 + (4 - 0)^2) = 3sqrt(2). The magnitude of BC is sqrt((0 - (-2))^2 + (0 - (-4))^2) = 2sqrt(5).

The dot product of BA and BC is:

BA · BC = (3sqrt(2))(-2) + (4)(-4) = -6sqrt(2) - 16

The magnitudes of BA and BC are:

|BA| = 3sqrt(2)
|BC| = 2sqrt(5)

Substituting into the dot product formula:

-6sqrt(2) - 16 = 3sqrt(2) * 2sqrt(5) * cos(theta)

Simplifying:

-6sqrt(2) - 16 = 6sqrt(10) * cos(theta)

cos(theta) = (-6sqrt(2) - 16)/(6sqrt(10)) = -(3sqrt(2) + 8)/3sqrt(10)

Using a calculator, we can find that cos(theta) ≈ -0.59846. Taking the inverse cosine, we have:

theta ≈ 126.87 degrees

So the size of angle ABC is approximately 126.87 degrees.

Sketch the graph for the following

A) (x²/9)_ (y²/25)=1
B) (y²/9)+(x²/4)=1
C) (x²/1)+(y²/1)=9

A) This is the equation of a hyperbola with center at the origin. The foci are at points (±sqrt(34)/3, 0), and the vertices are at points (±3, 0) and (0, ±5). Here's a sketch:

![hyperbola sketch](https://i.imgur.com/7QZqGZ4.png)

B) This is the equation of an ellipse with center at the origin. The foci are at points (0, ±sqrt(5)), and the vertices are at points (0, ±3) and (±2, 0). Here's a sketch:

![ellipse sketch](https://i.imgur.com/JPoJzRN.png)

C) This is the equation of a circle with center at the origin and radius 3. Here's a sketch:

![circle sketch](https://i.imgur.com/i5BYjA6.png)

find the derivatives of the following

A) y= (sinx/2x)
B) y= (2x² _ 5x)

A) To find the derivative of y = (sinx/2x), we use the quotient rule:

y' = [(2x)(cosx) - (sinx)(2)] / (2x)^2

Simplifying:

y' = (2cosx - 2sinx/x) / 4

y' = (1/2)(cosx - sinx/x)

So the derivative of y = (sinx/2x) is y' = (1/2)(cosx - sinx/x).

B) To find the derivative of y = (2x^2 - 5x), we use the power rule:

y' = 4x - 5

So the derivative of y = (2x^2 - 5x) is y' = 4x - 5.

sin 30= 0.5 and tan 60 will be