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Consider the family of circles passing through the two intersection…
Question
Consider the family of circles passing through the two intersection points of the circle x^2 + y^2 - 4x - 6y + 9 = 0 and the line x + y - 3 = 0. Exactly one member of this family passes through the origin. If the square of the radius of that circle equals m/n in lowest terms, find m+n.
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Step-by-step solution
The family of circles through the intersection points is S + lambda L = 0:
(x^2 + y^2 - 4x - 6y + 9) + lambda(x + y - 3) = 0.
(The given line genuinely cuts the circle at the real points (0,3) and (2,1), so the family is well-defined.)
Pass through the origin (0,0): 9 + lambda(-3) = 0 => lambda = 3.
Substitute lambda = 3:
x^2 + y^2 - 4x - 6y + 9 + 3x + 3y - 9 = 0 => x^2 + y^2 - x - 3y = 0.
Centre = (1/2, 3/2). Radius^2 = (1/2)^2 + (3/2)^2 - 0 = 1/4 + 9/4 = 10/4 = 5/2.
m/n = 5/2, m+n = 7.
Final answer7
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