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November 29th, 2009, 06:49 AM   #1
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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.
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November 29th, 2009, 07:59 AM   #2
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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.
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November 29th, 2009, 08:23 AM   #3
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Re:

Quote:
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.
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November 29th, 2009, 08:43 AM   #4
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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.
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November 29th, 2009, 04:46 PM   #5
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Re: circle

thanks again !
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November 30th, 2009, 05:10 AM   #6
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Re: circle

Quote:
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.
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November 30th, 2009, 09:20 AM   #7
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Symmetry considerations imply the fourth tangent has equation 3y + 4x = 12.
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November 30th, 2009, 04:37 PM   #8
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Re:

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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 ?
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File Type: jpg symmetry.jpg (4.7 KB, 68 views)
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December 1st, 2009, 09:06 AM   #9
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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.
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December 1st, 2009, 09:39 PM   #10
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Re:

Quote:
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 ?
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File Type: jpg symmetry 2.jpg (6.3 KB, 57 views)
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