My Math Forum Analytic Geometry- Circle

 Geometry Geometry Math Forum

 October 4th, 2012, 01:33 AM #1 Newbie   Joined: Oct 2012 Posts: 9 Thanks: 0 Analytic Geometry- Circle Show that the circles x² + y² - 4x + 2y - 20 =0 and 2x² + 2y² - 17x + 16y + 65 = 0 touch internally and find the point of contact. Another similar question : Prove that the circles x² + y² + 2x -8y + 8 = 0 and x² + y² +10x - 2y + 22 = 0 touch one another externally and find the point of contact. Please help. I do not understand what it means by the circles touching internally and externally. How to solve both this questions. What method should I use? Thanks
October 4th, 2012, 09:23 AM   #2
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Re: Analytic Geometry- Circle

1.) Here is a plot of the two circles, which shows what is meant by touching internally:

[attachment=0:ag4o32ow]internaltouch.jpg[/attachment:ag4o32ow]

To find the point where they touch, we may observe that this point must lie on the line passing through the centers of the two circles, so we may write the circles in the form:

$(x-h)^2+(y-k)^2=r^2$

$x^2+y^2-4x+2y-20=0$

$$$x^2-4x+4$$+$$y^2+2y+1$$=20+4+1$

$(x-2)^2+(y+1)^2=5^2$

We see that this circle is centered at $(2,-1)$

$2x^2+2y^2-17x+16y+65=0$

$2$$x^2-\frac{17}{2}x+\frac{289}{16}$$+2$$y^2+8y+16$$=-65+2$$\frac{289}{16}$$+2(16)$

$2$$x-\frac{17}{4}$$^2+2$$y+4$$^2=$$\frac{5}{2\sqrt{2}}\ )^2$ We see that this circle is centered at $\(\frac{17}{4},-4$$$

The slope of the line passing through the two centers is $m=\frac{3}{-\frac{9}{4}}=-\frac{4}{3}$

Using the point-slope formula, we find the line passing through the two centers is:

$y+1=-\frac{4}{3}(x-2)$

Now, using the equation of the first circle, we may substitute for $y+1$:

$(x-2)^2+$$-\frac{4}{3}(x-2)$$^2=5^2$

$\frac{25}{9}(x-2)^2=25$

$(x-2)^2=9$

We know the $x$ must be on the interval $$\frac{17}{4}-\frac{5}{2\sqrt{2}},\frac{17}{4}+\frac{5}{2\sqrt{2 }}$$ so we must take the positive root:

$x-2=3$

$x=5$

And so:

$y+1=-\frac{4}{3}(5-2)$

$y+1=-4$

$y=-5$

And so the point at which the two circles touch is $(5,-5)$.
Attached Images
 internaltouch.jpg (22.3 KB, 277 views)

 October 4th, 2012, 08:27 PM #3 Newbie   Joined: Oct 2012 Posts: 9 Thanks: 0 Re: Analytic Geometry- Circle Thank You for the clear explanation. For the circles to touch externally, all we have to do is find the gradient that cuts through the diameter of the two circles, is that right?
 October 4th, 2012, 08:29 PM #4 Newbie   Joined: Oct 2012 Posts: 9 Thanks: 0 Re: Analytic Geometry- Circle The way to solve both equations is the same? It doesn't matter whether it is internally or externally?
 October 4th, 2012, 08:52 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Analytic Geometry- Circle Yes, the method is the same. I neglected to show that the circles touch internally, which may be done by showing the distance between the centers of the circles is equal to the larger radius less the smaller one. For the second problem, you can show they touch externally by showing the distance between the centers is equal to the sum of the radii.
 October 4th, 2012, 10:48 PM #6 Newbie   Joined: Oct 2012 Posts: 9 Thanks: 0 Re: Analytic Geometry- Circle Thanks Mark!

 Tags analytic, circle, geometry

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# analytic geometry: the circle

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post condemath Geometry 1 August 29th, 2011 03:50 AM Rayne Geometry 0 November 2nd, 2010 06:32 AM Rayne Geometry 0 November 2nd, 2010 05:57 AM nortpron Geometry 6 April 5th, 2010 01:07 AM amkroes Geometry 12 November 7th, 2009 03:53 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top