# A circle has a center that falls on the line #y = 3/8x +3 # and passes through # ( 1 ,4 )# and #(2 ,9 )#. What is the equation of the circle?

The equation of the circle is

The circle's radius is

The circle's equation is

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The center of the circle can be found by determining the intersection point of the line y = (3/8)x + 3 and the perpendicular bisector of the line segment connecting the two given points (1, 4) and (2, 9).

The slope of the line connecting the two given points is (9 - 4)/(2 - 1) = 5. Since the perpendicular bisector will have a negative reciprocal slope, the slope of the perpendicular bisector is -1/5.

Using the midpoint formula, the coordinates of the midpoint of the line segment connecting (1, 4) and (2, 9) can be found as follows:

Midpoint (h, k) = ((1 + 2)/2, (4 + 9)/2) = (3/2, 13/2).

Now, we have the midpoint and the slope of the perpendicular bisector. Using the point-slope form of a line, the equation of the perpendicular bisector is y - (13/2) = -1/5(x - 3/2).

Simplifying, the equation becomes y = -1/5x + 13/2.

To find the intersection point of this line with y = (3/8)x + 3, we solve the system of equations:

y = -1/5x + 13/2 y = (3/8)x + 3

Solving for x and y yields:

-1/5x + 13/2 = 3/8x + 3 -1/5x - 3/8x = 3 - 13/2 -8/40x - 15/40x = 6 - 26/2 -23/40x = 6 - 13 -23/40x = -7

x = (-7)(-40/23) = 280/23

Substitute x = 280/23 into either equation to find y:

y = (3/8)(280/23) + 3 = 105/23 + 3 = (105 + 69)/23 = 174/23

Therefore, the center of the circle is (280/23, 174/23).

Now, the radius of the circle can be found using the distance formula between one of the given points and the center of the circle. Let's use (1, 4):

r = sqrt((280/23 - 1)^2 + (174/23 - 4)^2)

Finally, the equation of the circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius. Substituting the values, we get the equation of the circle.

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The equation of the circle is (x - 2)^2 + (y - 7)^2 = 13/2.

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To find the equation of the circle, we first need to determine its center and radius. Since the center of the circle falls on the line y = (3/8)x + 3, we can find it by solving for x and y when y = (3/8)x + 3.

Given that the coordinates of the center are (h, k), we can set up the equation as follows:

k = (3/8)h + 3

Now, we can find the center by substituting the coordinates of the given points into the equation y = (3/8)x + 3 to find the values of h and k.

For point (1, 4): 4 = (3/8)(1) + 3

For point (2, 9): 9 = (3/8)(2) + 3

Solving these equations, we find that the center of the circle is (1, 4).

Now, we can find the radius of the circle using the distance formula between the center and one of the given points. Let's use the point (1, 4):

r = √[(x2 - x1)^2 + (y2 - y1)^2] r = √[(1 - 1)^2 + (4 - 4)^2] r = √[0^2 + 0^2] r = 0

Since the radius is 0, the circle is actually just a point at (1, 4).

Therefore, the equation of the circle is x^2 + y^2 = 0.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- Two circles have the following equations: #(x +6 )^2+(y -5 )^2= 64 # and #(x -9 )^2+(y +4 )^2= 81 #. Does one circle contain the other? If not, what is the greatest possible distance between a point on one circle and another point on the other?
- A circle's center is at #(2 ,4 )# and it passes through #(1 ,2 )#. What is the length of an arc covering #(5pi ) /4 # radians on the circle?
- What is the equation of the circle with a center at #(5 ,-3 )# and a radius of #6 #?

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