A circle has a center that falls on the line #y = 2/7x +7 # and passes through # ( 3 ,4 )# and #(6 ,1 )#. What is the equation of the circle?

Answer 1

#(x - 63/5)^2 + (y - 53/5)^2 = ((3sqrt377)/5)^2#

The general Cartesian form for the equation of a circle is:

#(x-h)^2 + (y-k)^2 = r^2" [1]"#
Substitute the point #(3,4)# into equation [1]:
#(3-h)^2 + (4-k)^2 = r^2" [2]"#
Substitute the point #(6,1)# into equation [1]:
#(6-h)^2 + (1-k)^2 = r^2" [3]"#

Expand the squares of equation [2] and [3]:

#9-6h+h^2 + 16-8k+k^2 = r^2" [4]"# #36-12h+h^2 + 1-2k+k^2 = r^2" [5]"#

Subtract equation [5] from equation [4]:

#color(white)(....)9-6h+h^2 + 16-8k+k^2 = r^2" [4]"# #ul(-36+12h-h^2 - 1+2k-k^2 = -r^2)" [5]"# #-27+6h+0h^2+15-6k+0k^2= 0r^2#

Remove all of the 0 terms and move the constant to the right:

#6h - 6k = 12" [6]"#
Evaluate the given equation, #y = 2/7x +7#, at the center #(h,k)#:
#k = 2/7h+7" [7]"#

Substitute equation [7] into equation [6]:

#6h - 6(2/7h+7) = 12#

Solve for h:

#42h -12h- 294 = 84#
#30h = 378#
#h = 63/5#

Use equation [7] to find the value of k:

#k = 2/7(63/5)+7#
#k = 53/5#

Use equation [2] to find the value of r:

#(3-63/5)^2 + (4-53/5)^2 = r^2#
#(15/5-63/5)^2 + (20/5-53/5)^2 = r^2#
#(-48/5)^2 + (-33/5)^2 = 3393/25 = r^2#
#r = (3sqrt377)/5#

Use equation [1] to write the equation:

#(x - 63/5)^2 + (y - 53/5)^2 = ((3sqrt377)/5)^2#
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Answer 2

To find the equation of the circle, we first need to find the center of the circle, which lies on the line (y = \frac{2}{7}x + 7).

The center of the circle will have coordinates ((x_c, y_c)), where (x_c) and (y_c) satisfy the equation of the line. We can substitute the coordinates of one of the given points into the equation of the line to find the center.

Let's use the point ((3, 4)):

[4 = \frac{2}{7}(3) + 7]

Solving this equation gives (x_c = 3). Now, we can find (y_c):

[y_c = \frac{2}{7}(3) + 7]

[y_c = \frac{6}{7} + 7 = \frac{6}{7} + \frac{49}{7} = \frac{55}{7}]

So, the center of the circle is ((3, \frac{55}{7})).

Next, we need to find the radius of the circle, which is the distance between the center and one of the given points. Let's use the point ((6, 1)).

The distance formula between two points ((x_1, y_1)) and ((x_2, y_2)) is given by:

[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]

So, the radius (r) of the circle is:

[r = \sqrt{(6 - 3)^2 + (1 - \frac{55}{7})^2}]

[r = \sqrt{3^2 + \left(1 - \frac{55}{7}\right)^2}]

[r = \sqrt{9 + \left(1 - \frac{55}{7}\right)^2}]

[r = \sqrt{9 + \left(\frac{7}{7} - \frac{55}{7}\right)^2}]

[r = \sqrt{9 + \left(\frac{-48}{7}\right)^2}]

[r = \sqrt{9 + \frac{2304}{49}}]

[r = \sqrt{\frac{441 + 2304}{49}}]

[r = \sqrt{\frac{2745}{49}}]

[r = \frac{\sqrt{2745}}{7}]

Now, we have the center ((3, \frac{55}{7})) and the radius (r = \frac{\sqrt{2745}}{7}), so we can write the equation of the circle:

[(x - 3)^2 + \left(y - \frac{55}{7}\right)^2 = \left(\frac{\sqrt{2745}}{7}\right)^2]

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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.

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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