Question 413503

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Hi
Find the equation of the circle that has its center at (6,-5) and passes through (1,7).
(6,-5)
(1,7)
 r = sqrt(5^2 + (-12)^2) = 13      |D = {{{sqrt ((x1-x2)^2+(y1-y2)^2))}}}
Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}}
where Pt(h,k) is the center and r is the radius

   (x-6)^2 + (y+5)^2 = 169
{{{drawing(300,300,   -10,10,-10,10,  grid(1),
circle(6, -5,0.4),
circle(6, -5,13),
circle(1, 7,0.4),
graph( 300, 300, -10,10,-10,10))}}}