# A circle has a center that falls on the line #y = 12/7x +8 # and passes through # ( 9 ,1 )# and #(8 ,7 )#. What is the equation of the circle?

Equation of circle is

The circle equation's center-radius form is

formula (3), we obtain,

graph{(y-2)^2 + (x+3.5)^2 = 157.25 [-40, 40, -20, 20]} [Ans]

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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.

First, find the center of the circle by finding the intersection point of the line and the perpendicular bisector of the segment joining the given points.

The midpoint of the segment joining (9, 1) and (8, 7) is: ( \left(\frac{9 + 8}{2}, \frac{1 + 7}{2}\right) = (8.5, 4) )

The slope of the line ( y = \frac{12}{7}x + 8 ) is ( m = \frac{12}{7} ), so the slope of the perpendicular bisector is ( m_{\text{perpendicular}} = -\frac{7}{12} ).

Using the point-slope form of a line, the equation of the perpendicular bisector passing through (8.5, 4) is: ( y - 4 = -\frac{7}{12}(x - 8.5) )

Now, find the intersection of this line with the given line ( y = \frac{12}{7}x + 8 ) to get the center of the circle.

Solving these two equations simultaneously will give the coordinates of the center.

After finding the center, use either given point to find the radius.

Finally, substitute the values of the center and radius into the equation of the circle.

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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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