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: 463 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 $(6,3)$. $\displaystyle (3h)^2+(2k)^2=r^2$ $\displaystyle (6h)^2+(3k)^2=r^2$ Equating the squares of the radii, we obtain: $\displaystyle (3h)^2+(2k)^2=(6h)^2+(3k)^2$ $\displaystyle 96h+h^2+44k+k^2=3612h+h^2+96k+h^2$ $\displaystyle 3h+k=16$ 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: $\displaystyle (xh)^2+(y+3h16)^2=r^2$ 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: $\displaystyle (22yh)^2+(y+3h16)^2=r^2$ Expanding and rearranging gives us the following quadratic in $y$: $\displaystyle 5y^2+(10h40)y+(10h^2100h+260r^2)=0$ Since the circles are tangent to the line, we require the discriminant to be zero: $\displaystyle (10h40)^24(5)(10h^2100h+260r^2)=0$ From this, we obtain: $\displaystyle r^2=5(h6)^2$ And so, substituting for $r^2$ into either of our first 2 equations (I'll choose the first), we obtain $\displaystyle (3h)^2+(216+3h)^2=5(h6)^2$ $\displaystyle h^26h+5=0$ $\displaystyle (h1)(h5)=0$ For $h=1$ we have: $\displaystyle (x1)^2+(y13)^2=125$ For $h=5$ we have: $\displaystyle (x5)^2+(y1)^2=5$ So, it turns out, that I would work this problem essentially the same way you did. 
March 4th, 2017, 02:33 PM  #3  
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  Quote:
 

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circles, equations, finding, geometry 
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