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 March 4th, 2017, 02:52 AM #1 Newbie   Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus Finding Equations of Two Circles I want to ask if there any easier method for solving below questions. The proofing already answered but the second question puzzle me a bit. It quite tedious for solving three equations. The answer is correct but the working steps quite long. I skipped for showing the final working step. Anybodies have different method?
 March 4th, 2017, 09:58 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs I think the way I would work this is to first find the family of circles passing through the points $(3,2)$ and $(6,3)$. $\displaystyle (3-h)^2+(2-k)^2=r^2$ $\displaystyle (6-h)^2+(3-k)^2=r^2$ Equating the squares of the radii, we obtain: $\displaystyle (3-h)^2+(2-k)^2=(6-h)^2+(3-k)^2$ $\displaystyle 9-6h+h^2+4-4k+k^2=36-12h+h^2+9-6k+h^2$ $\displaystyle 3h+k=16$ This implies that the centers of the circles do in fact lie along the line: $\displaystyle 3x+y=16$ And so, our family of circles can be written as: $\displaystyle (x-h)^2+(y+3h-16)^2=r^2$ To find the members of this family tangent to the line $\displaystyle x+2y=2$, we may use this line to substitute for $x$ into our family of circles: $\displaystyle (2-2y-h)^2+(y+3h-16)^2=r^2$ Expanding and rearranging gives us the following quadratic in $y$: $\displaystyle 5y^2+(10h-40)y+(10h^2-100h+260-r^2)=0$ Since the circles are tangent to the line, we require the discriminant to be zero: $\displaystyle (10h-40)^2-4(5)(10h^2-100h+260-r^2)=0$ From this, we obtain: $\displaystyle r^2=5(h-6)^2$ And so, substituting for $r^2$ into either of our first 2 equations (I'll choose the first), we obtain $\displaystyle (3-h)^2+(2-16+3h)^2=5(h-6)^2$ $\displaystyle h^2-6h+5=0$ $\displaystyle (h-1)(h-5)=0$ For $h=1$ we have: $\displaystyle (x-1)^2+(y-13)^2=125$ For $h=5$ we have: $\displaystyle (x-5)^2+(y-1)^2=5$ So, it turns out, that I would work this problem essentially the same way you did. Thanks from greg1313 and palongze
March 4th, 2017, 02:33 PM   #3
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Joined: Jan 2017
From: Malaysia

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Math Focus: calculus
Quote:
 Originally Posted by MarkFL I think the way I would work this is to first find the family of circles passing through the points $(3,2)$ and $(6,3)$. $\displaystyle (3-h)^2+(2-k)^2=r^2$ $\displaystyle (6-h)^2+(3-k)^2=r^2$ Equating the squares of the radii, we obtain: $\displaystyle (3-h)^2+(2-k)^2=(6-h)^2+(3-k)^2$ $\displaystyle 9-6h+h^2+4-4k+k^2=36-12h+h^2+9-6k+h^2$ $\displaystyle 3h+k=16$ This implies that the centers of the circles do in fact lie along the line: $\displaystyle 3x+y=16$ And so, our family of circles can be written as: $\displaystyle (x-h)^2+(y+3h-16)^2=r^2$ To find the members of this family tangent to the line $\displaystyle x+2y=2$, we may use this line to substitute for $x$ into our family of circles: $\displaystyle (2-2y-h)^2+(y+3h-16)^2=r^2$ Expanding and rearranging gives us the following quadratic in $y$: $\displaystyle 5y^2+(10h-40)y+(10h^2-100h+260-r^2)=0$ Since the circles are tangent to the line, we require the discriminant to be zero: $\displaystyle (10h-40)^2-4(5)(10h^2-100h+260-r^2)=0$ From this, we obtain: $\displaystyle r^2=5(h-6)^2$ And so, substituting for $r^2$ into either of our first 2 equations (I'll choose the first), we obtain $\displaystyle (3-h)^2+(2-16+3h)^2=5(h-6)^2$ $\displaystyle h^2-6h+5=0$ $\displaystyle (h-1)(h-5)=0$ For $h=1$ we have: $\displaystyle (x-1)^2+(y-13)^2=125$ For $h=5$ we have: $\displaystyle (x-5)^2+(y-1)^2=5$ So, it turns out, that I would work this problem essentially the same way you did.
Nice one. I think this questions is a bit tedious to solve. Anyway Thank you.

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