My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Thanks Tree3Thanks
  • 2 Post By MarkFL
  • 1 Post By palongze
Reply
 
LinkBack Thread Tools Display Modes
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?

palongze is offline  
 
March 4th, 2017, 09:58 AM   #2
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,155
Thanks: 462

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
MarkFL is offline  
March 4th, 2017, 02:33 PM   #3
Newbie
 
Joined: Jan 2017
From: Malaysia

Posts: 20
Thanks: 3

Math Focus: calculus
Quote:
Originally Posted by MarkFL View Post
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.
Thanks from MarkFL
palongze is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
circles, equations, finding, geometry



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Finding distances within circles u0362565 Algebra 4 December 16th, 2013 03:22 PM
Circles from equations alexpasty Algebra 3 April 17th, 2013 12:33 PM
Find The Standard Equations Of The Circles... soulrain Algebra 8 January 8th, 2012 07:01 PM
re finding the area of circles with a regular octagon ibros06 Algebra 3 June 13th, 2010 06:50 PM
Equations related to the circles in a Cartesian graph. johnny Algebra 1 April 26th, 2007 06:48 PM





Copyright © 2017 My Math Forum. All rights reserved.