A circle has a center that falls on the line #y = 1/7x +4 # and passes through # ( 7 ,8 )# and #(3 ,6 )#. What is the equation of the circle?

Answer 1

#color(blue)((x-91/15)^2+(y-73/15)^2=481/45)#

The equation of a circle is given by:

#(x-h)^2+(y-k)^2=r^2#
Where #(h,k)# are the #x and y# coordinates of the centre respectively and #r# is the radius.
The centre lie on the line #y=1/7x+4#
#:.#
#k=1/7h+4 \ \ \ \ \ \ \[1]#
#(7,8) and (3,6)# lie on the circle:
#(7-h)^2+(8-k)^2=r^2 \ \ \ \ \ \ \[2]#
#(3-h)^2+(6-k)^2=r^2 \ \ \ \ \ \ \[3]#
Subtract #[3]# from #[2]#
#68-8h-4k=0#
Substituting #k=1/7h+4#
#68-8h-4(1/7h+4)=0#
#52-60/7h=0=>h=91/15#
Substituting in #[1]#:
#k=1/7(91/15)+4#
#k=73/15#
We now need the value of #r^2#
Using coordinate #(3,6)#
#(3-91/15)^2+(6-73/15)^2=r^2#
#r^2=481/45#

So the equation of the circle is:

#color(blue)((x-91/15)^2+(y-73/15)^2=481/45)#
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Answer 2

The equation of a circle with center ((h, k)) and radius (r) is given by ((x - h)^2 + (y - k)^2 = r^2).

First, we need to find the center of the circle. Since the center lies on the line (y = \frac{1}{7}x + 4), we can substitute this equation into the general form of the circle equation to find the center. So, we have:

((x - h)^2 + (\frac{1}{7}x + 4 - k)^2 = r^2).

Next, we use the fact that the circle passes through the points ((7, 8)) and ((3, 6)). Plugging these points into the circle equation, we get two equations:

((7 - h)^2 + (\frac{1}{7} \cdot 7 + 4 - k)^2 = r^2), and

((3 - h)^2 + (\frac{1}{7} \cdot 3 + 4 - k)^2 = r^2).

Now, we have a system of three equations with three unknowns ((h), (k), and (r^2)). Solving this system will give us the values of (h), (k), and (r^2), which we can then use to write the equation of the circle.

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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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