Question 1144303
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The circle touches the line x+y=8 at (2,6); i.e., it is tangent to that line at that point.  That means the center of the circle is on the line that is perpendicular to the given line at that point.<br>
The given line has a slope of -1, so the line containing that radius has a slope of 1.  Since it passes through (2,6), the equation of the line containing that radius is y=x+4.<br>
So let an arbitrary point on the line containing the center of the circle be (a,a+4).  We need the distances from (a,a+4) to (2,6) and from (a,a+4) to (4,0) to be the same.<br>
It is easier to solve the problem using the squares of the distances....<br>
square of the distance from (a,a+4) to (2,6): {{{(a-2)^2+(a-2)^2}}}
square of the distance from (a,a+4) to (4,0): {{{(a-4)^2+(a+4)^2}}}<br>
Then<br>
{{{(a-2)^2+(a-2)^2 = (a-4)^2+(a+4)^2}}}
{{{2(a^2-4a+4) = (a^2-8a+16)+(a^2+8a+16)}}}
{{{2a^2-8a+8 = 2a^2+32}}}
{{{-8a = 24}}}
{{{a = -3}}}<br>
The center of the circle is (a,a+4) = (-3,1).<br>
CHECK:
distance from (-3,1) to (2,6) = {{{sqrt(5^2+5^2) = sqrt(50)}}}
distance from (-3,1) to (4,0) = {{{sqrt(7^2+1^2) = sqrt(50)}}}