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:
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:
C) This is the equation of a circle with center at the origin and radius 3. Here's a sketch:
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
To find the value of tan 60, we can use the identity:
tan(x) = sin(x) / cos(x)
and substitute x = 60 degrees (which is equivalent to pi/3 radians).
We know that sin 60 = sqrt(3)/2, but we still need to find cos 60.
Using the identity:
sin^2(x) + cos^2(x) = 1
We can solve for cos 60:
(sin 60)^2 + (cos 60)^2 = 1
(3/4) + (cos 60)^2 = 1
(cos 60)^2 = 1 - 3/4
(cos 60)^2 = 1/4
cos 60 = ±1/2
Since the cosine of 60 degrees is positive (since it falls in the first quadrant), we take the positive root:
cos 60 = 1/2
Now we can substitute into the formula for tangent:
tan 60 = sin 60 / cos 60
tan 60 = (sqrt(3)/2) / (1/2)
tan 60 = sqrt(3)
So the value of tan 60 is sqrt(3) or approximately 1.732.
the end coordinates of a line segment are A(-5,1) and B(3,1). A line from point C(-1,4) intersects AB at D. Line CD is perpendicular to the line AB.
A) the point of segment AB
B) determine the gradient of segment AB
C) what is the length of line CD
D) the equation of line CD
E) if AB is the diameter of the circle, then find the equation of the circle
F)the area of the shape created by coordinates ABD
G) what is the angle ACD
A) Since the line CD intersects AB at D, we know that D lies on AB. The x-coordinate of D is the average of the x-coordinates of A and B, and the y-coordinate of D is given by the y-coordinate of A or B since AB is a horizontal line:
x-coordinate of D = (-5 + 3)/2 = -1
y-coordinate of D = 1
So the point of intersection is D(-1,1).
B) The gradient of AB can be found using the formula:
m = (y2 - y1)/(x2 - x1)
where (x1,y1) and (x2,y2) are the coordinates of A and B, respectively:
m = (1 - 1)/(3 - (-5)) = 0/8 = 0
So the gradient of AB is 0.
C) Since line CD is perpendicular to AB, it is parallel to the y-axis and has length equal to the x-coordinate of D minus the x-coordinate of C.
Length of CD = D_x - C_x = (-1) - (-1) = 2
So the length of CD is 2.
D) Since line CD is parallel to the y-axis, its equation is of the form:
x = k
where k is a constant. To find k, we know that the line passes through (x,y) = (-1,4):
x = -1
So the equation of line CD is x = -1.
E) If AB is the diameter of the circle, then the center of the circle is the midpoint of AB, which is ((-5+3)/2, (1+1)/2) = (-1,1). The radius is half the length of AB, which is 4. Therefore, the equation of the circle is:
(x - (-1))^2 + (y - 1)^2 = 4^2
(x + 1)^2 + (y - 1)^2 = 16
F) To find the area of triangle ABD, we can use the formula:
Area = (1/2) * base * height
where the base is the distance between A and B, and the height is the distance between AB and CD:
Base AB = 3 - (-5) = 8
Height CD = 4 - 1 = 3
Area ABD = (1/2) * 8 * 3 = 12
So the area of the shape created by coordinates ABD is 12.
G) Angle ACD is the complement of angle ACB. The slope of line AB is zero, so line CD is vertical and the slope of line AC is undefined. Therefore, angle ACB is a right angle. So angle ACD is also a right angle (since it is the complement).
which of the following is the major segment of the elipse vertical on the grid
A) (x²/a²)_ (y²/b²)=1
B) (x²/a²)+(y²/b²)=1
C) (x²/b²)_(y²/a²)=1
D) (x²/b²)+(y²/a²)=1
If the major axis of an ellipse is vertical, then the larger denominator (between a² and b²) will correspond to the y-term of the equation.
In options A and B, the larger denominator is b², so the major axis is vertical for both of these options.
In options C and D, the larger denominator is a², so the minor axis is vertical for both of these options.
Therefore, the answer is either A or B.
what type of equation is this ((x_2²)/8)+((y_2²)/12.5)=2
what type of equation is this ((x_2²)/8)+((y_2²)/12.5)=2
Hyperbola, elipse, circle or parabola