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 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: 20,757 Thanks: 2138 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
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 Originally Posted by skipjack 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.
Thanks a lot, you are right, x and y > 0 . But I have a question here: why are you so sure that the x and y intercepts are (r,0) and (0,r) respectively? I mean is the centre of the circle aligned to the y and x intercepts?

Last edited by skipjack; November 5th, 2014 at 09:19 AM. November 29th, 2009, 08:43 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 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
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 Originally Posted by mikeportnoy 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.
This is the continuation of the question which I am unsure of:

Find the equation of the fourth tangent to the two circles. November 30th, 2009, 09:20 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 Symmetry considerations imply the fourth tangent has equation 3y + 4x = 12. November 30th, 2009, 04:37 PM   #8
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 Originally Posted by skipjack Symmetry considerations imply the fourth tangent has equation 3y + 4x = 12.
Thanks again skipjack , is it gonna look sth like this (attachment ) ? How did you obtain the other equation , just by switching the x and y ? Or its sort of a transformation ?
Attached Images symmetry.jpg (4.7 KB, 68 views) December 1st, 2009, 09:06 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 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
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 Originally Posted by skipjack 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.
ok , can u check my diagram again ..

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 ?
Attached Images symmetry 2.jpg (6.3 KB, 57 views) Tags circle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post yeoky Algebra 4 May 3rd, 2014 01:06 AM Aska123 Algebra 9 January 29th, 2014 11:06 PM gailplush Algebra 5 August 8th, 2010 04:19 AM sir anrava Algebra 1 January 19th, 2010 11:18 AM Titan Algebra 0 March 7th, 2008 05:14 PM

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