November 29th, 2009, 06:49 AM  #1 
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  circle
A circle, radius r, with its centre in the positive quadrant (x<0 , y<0) touches both the x and y axes. The circle also touches the line 3x+4y=12. Find the two values of r and obtain the equations of the two corresponding circles.

November 29th, 2009, 07:59 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,372 Thanks: 2009 
Did you mean "positive quadrant (x>0 , y>0)" rather than "positive quadrant (x<0 , y<0)? If so, the circle has equation (x  r)² + (y  r)² = r², where r is its radius. Since the circle touches the line 3x + 4y = 12, i.e., y = 3  (3/4)x, using that equation to substitute for y in the equation of the circle results in a quadratic equation in x that has to have equal roots. If the quadratic equation is written in the form ax² + bx + c = 0, the coefficients a, b, and c must therefore satisfy the equation b²  4ac = 0, and that will be a quadratic equation in r that you can solve to find the two value of r that the question asks for. Last edited by skipjack; November 5th, 2014 at 09:18 AM. 
November 29th, 2009, 08:23 AM  #3  
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  Re: Quote:
Last edited by skipjack; November 5th, 2014 at 09:19 AM.  
November 29th, 2009, 08:43 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,372 Thanks: 2009 
The circle's centre is in the first quadrant and is at the same distance r, where r is the radius, from the two axes (since the circle touches them), so it's the point (r, r). There's a standard formula, d = Am + Bn + C/√(A² + B²), for the distance of a point (m, n) from the line with equation Ax + By + C = 0. If you are allowed to use that formula, you know that the circle's centre is at a distance r from the line 3x + 4y  12 = 0, so you can find r by solving r = 3r + 4r  12/√(3² + 4²). Squaring both sides of that equation gives you a quadratic equation in r that factorizes easily (or you can just solve the linear equations r = ±(3r + 4r  12)/√(3² + 4²)). Last edited by skipjack; November 5th, 2014 at 09:23 AM. 
November 29th, 2009, 04:46 PM  #5 
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  Re: circle
thanks again !

November 30th, 2009, 05:10 AM  #6  
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  Re: circle Quote:
Find the equation of the fourth tangent to the two circles.  
November 30th, 2009, 09:20 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,372 Thanks: 2009 
Symmetry considerations imply the fourth tangent has equation 3y + 4x = 12.

November 30th, 2009, 04:37 PM  #8  
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  Re: Quote:
 
December 1st, 2009, 09:06 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,372 Thanks: 2009 
Yes, just by switching the x and y, which implements reflection in the line y = x (the line through the centres of the circles). Your diagram isn't correct.

December 1st, 2009, 09:39 PM  #10  
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  Re: Quote:
it doesn't seem to be correct , since the gradient of both tangents aer negative .. shouldn't it look like sth in my previous diagram ?  

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