Question 1050916
Find the equation of the circle of radius sqrt(29) tangent to the line 2x-5y+16=0 and passing through (9,-3)
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There are 2 circles that fit.
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Draw a circle of radius sqrt(29) with its center at (9,-3), call it circle Q.
The centers of the 2 circles with be on circle Q.
The centers' distance from the line will be sqrt(29).
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For the distance from a line:
d = |A*p + B*q + C|/sqrt(A^2 + B^2) where the line is Ax + By + C = 0 and (p,q) is the point.
d = |2*p - 5*q + 16|/sqrt(p^2 + q^2) = sqrt(29)
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That's too complicated.
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Find a line parallel to the given line and sqrt(29) from it.
Slope of the given line is 2/5 and the y-int is 16/5
The slope 2/5 is the tangent of the angle between the line and the x-axis.
Call the difference in the y-ints z.
cos(atan(2/5)) = sqrt(29)/z
z = 5.8
y-int of the parallel line = 16/5 - 5.8 = -13/5
Parallel line is y = (2/5)x - 13/5
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Find the 2 intersections of the line and circle Q.
Circle Q: (x-9)^2 + (y+3)^2 = 29
Sub for y
(x-9)^2 + ((2/5)x - 13/5 +3)^2 = 29
(x-9)^2 + ((2/5)x + 2/5)^2 = 29
Multiply by 25 to eliminate fractions.
(5x-45)^2 + (2x + 2)^2 = 725
25x^2 - 450x + 2025 + 4x^2 + 8x + 4 = 725
29x^2 - 442x + 1304 = 0
*[invoke solve_quadratic_equation 29,-442,1304]
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x = 4; y = -1  --> (x-4)^2 + (y+1)^2 = 29 is one circle.
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x = 326/29; y = 55/29 --> (x - 326/29)^2 + (y - 55/29)^2 = 29
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