Question 835835
Find an equation(s) of the circle(s) tangent to 2x - 3y + 6 = 0 at (3,4); center on 3x + 2y - 17 = 0
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The 2 lines intersect at (3,4) --> an infinite # of circles that fit.
3x + 2y - 17 = 0
y = -3x/2 + 17/2
Pick any point on the line and find its distance from (3,4)
Use (h,k) for the point
{{{d^2 = (k-4)^2 + (h-3)^2}}}
d is the radius
{{{(x-h)^2 + (y-k)^2 = d^2}}}
{{{x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = (k-4)^2 + (h-3)^2}}}
{{{x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = k^2-8k+16 + h^2-6h+9}}}
{{{x^2 - 2hx + y^2 - 2ky = -8k - 6h + 25}}}
{{{x^2 - h(2x - 6) + y^2 - k(2y - 8) = 25}}}
y = -3x/2 + 17/2 --> k = -3h/2 + 17/2
{{{x^2 - h(2x + 6) + y^2 + (3h-17)(y - 4) = 25}}}
{{{x^2 - h(2x + 6) + y^2 + 3hy - 12h - 17y + 68 = 25}}}
{{{x^2 - h(2x - 3y + 6) + y^2 - 17y + 43 = 0}}}
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Pick any value for h and you get a circle that fits.