Question 912053
a circle touches the circle (x+1)^2+y^2=9 and passes through (1,0). find the locus of its center. PLEASE GIVE SOLUTION (ANSWER: 20x^2+36y^2=45).thx XD
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Do you mean it's tangent to the circle?
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(1,0) is inside the given circle, so the circles will be internally tangent.
The radius of (x+1)^2+y^2=9 is 3.
The center of the circles is (h,k)
The distance from (h,k) to (1,0) = {{{sqrt(k^2 + (h-1)^2)}}}
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The distance also equals 3 - [the distance from (h,k) to (-1,0)]
= {{{3 - sqrt((h+1)^2 + k^2)}}}
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{{{sqrt((h+1)^2 + k^2) = 3 - sqrt(k^2 + (h-1)^2)}}}
{{{(h+1)^2 + k^2 = 9 - 6sqrt(k^2 + (h-1)^2) + k^2 + (h-1)^2}}}
{{{h^2 + 2h + 1 + k^2 = 9 - 6sqrt(k^2 + (h-1)^2) + k^2 + h^2 - 2h + 1}}}
{{{4h - 9 = -6sqrt(k^2 + (h-1)^2)}}}
{{{16h^2 - 72h + 81 = 36*(k^2 + h^2 - 2h + 1)}}}
{{{16h^2 - 72h + 81 = 36k^2 + 36h^2 - 72h + 36)}}}
{{{16h^2 + 45 = 36k^2 + 36h^2)}}}
{{{ 45 = 36k^2 + 20h^2)}}}
{{{36k^2 + 20h^2 - 45 = 0)}}}
Sub x & y for (h,k)
{{{20x^2 + 36y^2 = 45}}}