A circle has a center that falls on the line #y = 3/8x +8 # and passes through # ( 7 ,4 )# and #(2 ,9 )#. What is the equation of the circle?

Answer 1

#(x+48/5)^2+(y+58/5)^2= 1613/25#
See below for geometric representation

Given : The center of the a circle lies on the equation of line #y=3/8x+8# and passes through points #A(a_x, a_y) = A(7,4)# and #B(b_x, b_y)=B(2,9)# Required : Equation of the circle Solution strategy: a) Equation of the circle centered at #O(x_c, y_c)# #=> (x+x_c)^2 + (y+y_c)^2= r^2# The center of the circle #(x_c, y_c)=(x, 3/8x+8)# b) From the distance formula: #r^2 = (x-x_c)^2 + (y-y_c)^2# such that #A=>x: x= a_x=7 and y: y=a_y=4# #B=>x: x= b_x=2 and y: y=b_y=9#
We start with b) and writing two Distance Formula equations for A and B respectively: #A=> r^2= (x-7)^2+(3/8x+8-4)^2# # =(x-7)^2+(3/8x+4)^2#
#B=> r^2= (x-2)^2+(3/8x+8-9)^2# # = (x-2)^2+(3/8x-1)^2#
Now both #A and B# lie on the circle so thy are equal to one another, i.e #r^2=r^2# thus, # (x-7)^2+(3/8x+4)^2 = (x-2)^2+(3/8x-1)^2# Rearrange and clean up #64(x-7)^2 + (3x+32)^2 = 64(x-2)^2 + (3x-8)^2# Expand and eliminate the square in #x# terms: #64(x^2-14x+49) + (9x^2+192x+1024) = # #64(x^2-4x+4) + (9x^2-48x+64)# #cancel(73x^2)-704x+4160=cancel(73x^2)-304x+320# Solve for #x# #x=48/5#. Now insert #x=48/5# on the equation of line #y=3/8x+8# #y=3/8*48/5 +8 = 58/5# Thus the center, #O(x_c, y_c)= O(48/5, 58/5)#
To verify calculate the radiuses, #bar(OA) and bar(OB)# #(bar(OA))^2=(x_c-7)^2+(y_c-4)^2=(48/5-7)^2+(58/5-4)^2# #(bar(OA))^2 = 169/25 + 1444/25# #(bar(OB))^2=(x_c-2)^2+(y_c-9)^2=(48/5-2)^2+(58/5-9)^2# #(bar(OB))^2 = 1444/25 + 169/25# #(bar(OA))^2=(bar(OB))^2 = r^2#
Now from strategy a) we have #(x+x_c)^2+(y+y_c)^2= r^2# replace #(x_c, y_c) and r^2# by #(48/5, 58/5) and r^2=1613/25# #(x+48/5)^2+(y+58/5)^2= 1613/25#
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Answer 2

The equation of the circle can be found using the formula for the standard form of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

First, we need to find the center of the circle. Since the center falls on the line y = (3/8)x + 8, we can set this equation equal to y and solve for x: y = (3/8)x + 8 Substitute y with 4 (from the point (7, 4)): 4 = (3/8)(7) + 8 Then, solve for x: x = 8

So, the x-coordinate of the center of the circle is 8.

Now, let's find the y-coordinate of the center by substituting x with 8 in the equation of the line: y = (3/8)(8) + 8 Simplify: y = 3 + 8 y = 11

So, the y-coordinate of the center of the circle is 11.

Now, we need to find the radius of the circle. We can use the distance formula between the center of the circle (h, k) and one of the given points on the circle to find the radius.

Using the point (7, 4): r^2 = (x2 - x1)^2 + (y2 - y1)^2 r^2 = (7 - 8)^2 + (4 - 11)^2 r^2 = (-1)^2 + (-7)^2 r^2 = 1 + 49 r^2 = 50

Now we have the center (h, k) = (8, 11) and the radius squared r^2 = 50. We can plug these values into the standard form of the circle equation: (x - 8)^2 + (y - 11)^2 = 50

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Answer from HIX Tutor

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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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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