Question 1081946
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Find equation when the center on the line x - 2y - 9 = 0 and passes through the points (7,-2) and (5,0). 
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<pre>
The center of the circle lies on the perpendicular bisector to the segment connecting the points (7,-2) and (5,0).

The midpoint is (6,-1).     (6 = {{{(7+5)/2}}},  -1 = {{{(-2+0)/2}}})


The segment connecting the given points has the slope {{{(0-(-2))/(5-7)}}} = {{{2/(-2)}}} = -1.

Therefore, the perpendicular bisector has the slope 1.

Hence, the perpendicular bisector passing through the point (6,-1) has the equation

y - (-1) = 1*(x-6),   or,  which is the same,  y+1 = x - 6.


Hence, the center of the circle lies at the intersection of these two lines

x - 2y - 9 = 0    (1)   and
y+1 = x - 6.      (2)

This system is the same as

x - 2y = 9,       (3)
x -  y = 7.       (4)

To solve the system (and to find the intersection point) subtract (4) from (3) both sides). You will get

-y = 2,   or   y = -2.

Then x = 7 + y = 7 + (-2) = 5   (from (4).

Thus the point (5,-2) is the center.


The distance from the center (5,-2) to the given point (5,0) is {{{sqrt((5-5)^2 + (0-(-2)^2))}}} = {{{sqrt(0^2 + 2^2)}}} = {{{sqrt(4)}}} = 2.


Then the equation of the circle is 

{{{(x-5)^2 + (y-(-2))^2}}} = 4,   or

{{{(x-5)^2 + (y+2)^2}}} = 4.
</pre>

Solved.


{{{
          drawing( 400, 400, -2.5, 8.5, -7.5, 3.5, 
          grid(1), 

      red(circle(5, -2, 2)), 
          circle(5, -2, 0.1), 

          circle (7, -2, 0.18),
          circle (5, 0, 0.18),

          graph( 400, 400, -2.5, 8.5, -7.5, 3.5,
          -3x+13)
)}}}



The solution by "josgarithmetic" is WRONG.