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#### Albert

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- Jan 25, 2013

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- Thread starter Albert
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- Jan 25, 2013

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- Jan 26, 2012

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My solution

This is going to be algebraic. I'm assuming someone will give a geometric answer. If we flip the picture, then we can come up with the equation of the second circle (the one tangent to the line)

$ (x-3s)^2+(y-s)^2 = s^2$

and the equation of the straight line itself $ y = 9 - \dfrac{9}{12}x$

Now if $(x_0,y_0)$ is the point on the circle whose tangent is the line then we can also come up with the tangent line using calculus, namely

$y - y_0 = - \dfrac{(x_0 - 3s)}{(y_0-s)}( x - x_0)$

As the two lines must be the same gives (eliminate $y$ and isolate the coefficients wrt $x$)

$\dfrac{3}{4}-{\dfrac {{ x_0}}{{ y_0}-s}}+3\,{\dfrac {s}{{ y_0}-s}}=0\;\;\;(1)$

$-9+{\it y_0}+{\dfrac {{{ x_0}}^{2}}{{ y_0}-s}}-3\,{\dfrac {s{ x_0}}

{{\it y_0}-s}}

= 0\;\;\;\;(2)$

Furthermore, we know that $(x_0,y_0)$ is on the circle so

$ (x_0-3s)^2+(y_0-s)^2 = s^2\;\;\;(3)$

From (1) we find that $y_0 = -3s + \dfrac{4}{3} x_0$ and with this, from (2) we obtain

$x_0 = \dfrac{36}{25} s+\dfrac{108}{25}$ and with these two gives (3) as

${\frac {72}{25}}\, \left( s-2 \right) \left( 2\,s-9 \right) =0$

Clearly $s = 9/2$ is too large thus giving $s = 2$, the radius of the circle.

$ (x-3s)^2+(y-s)^2 = s^2$

and the equation of the straight line itself $ y = 9 - \dfrac{9}{12}x$

Now if $(x_0,y_0)$ is the point on the circle whose tangent is the line then we can also come up with the tangent line using calculus, namely

$y - y_0 = - \dfrac{(x_0 - 3s)}{(y_0-s)}( x - x_0)$

As the two lines must be the same gives (eliminate $y$ and isolate the coefficients wrt $x$)

$\dfrac{3}{4}-{\dfrac {{ x_0}}{{ y_0}-s}}+3\,{\dfrac {s}{{ y_0}-s}}=0\;\;\;(1)$

$-9+{\it y_0}+{\dfrac {{{ x_0}}^{2}}{{ y_0}-s}}-3\,{\dfrac {s{ x_0}}

{{\it y_0}-s}}

= 0\;\;\;\;(2)$

Furthermore, we know that $(x_0,y_0)$ is on the circle so

$ (x_0-3s)^2+(y_0-s)^2 = s^2\;\;\;(3)$

From (1) we find that $y_0 = -3s + \dfrac{4}{3} x_0$ and with this, from (2) we obtain

$x_0 = \dfrac{36}{25} s+\dfrac{108}{25}$ and with these two gives (3) as

${\frac {72}{25}}\, \left( s-2 \right) \left( 2\,s-9 \right) =0$

Clearly $s = 9/2$ is too large thus giving $s = 2$, the radius of the circle.

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- Feb 7, 2012

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- #4

- Jan 25, 2013

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I will use geometry ,

maybe someone want to try it first

maybe someone want to try it first

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- #5

- Feb 7, 2012

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I will use geometry ,

maybe someone want to try it first

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- #6

- Jan 25, 2013

- 1,225

yes ,you got it