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Math Help - Equation of Circle Tangent to Y-Axis, X-Axis and Line

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    Equation of Circle Tangent to Y-Axis, X-Axis and Line

    Please help me with the process of finding the equation of a circle in the first quadrant that is tangent to the x-axis, y-axis, and a line of the form ax+by+c=0. Thanks.
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    Quote Originally Posted by juankim111222 View Post
    Please help me with the process of finding the equation of a circle in the first quadrant that is tangent to the x-axis, y-axis, and a line of the form ax+by+c=0. Thanks.

    A circle (x-a)^2+(y-b)^2=r^2 is tangent to the x-axis iff |b|=r , is tangent to the y-axis iff |a|=r , and since the circle's center is in the first quadrant this means that the circle's equation is (x-r)^2+(y-r)^2=r^2 , and center at (r,r)

    Finally, the distance from a given point (x_1,y_1) to a given line Ax+By+C=0 is given by \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} . Do some maths now.

    Tonio
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    Thanks. So is there any way to find the center (r,r) if the only information given is the tangent line 3x+4y-12=0?
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    Quote Originally Posted by juankim111222 View Post
    Thanks. So is there any way to find the center (r,r) if the only information given is the tangent line 3x+4y-12=0?
    Just find where the line crosses the x and y axes.

    y=0\ \Rightarrow\ 3x-12=0\ \Rightarrow\ x=4

    x=0\ \Rightarrow\ 4y-12=0\ \Rightarrow\ y=3

    Then the triangle area is 2(3)=6

    and the hypotenuse from Pythagoras' theorem is 5.

    Then the radius is

    r=\frac{2A}{sum\ of\ the\ side\ lengths}=\frac{2(6)}{5+4+3}=1

    Hence, the centre is (1,1) and the radius is 1.
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    Quote Originally Posted by juankim111222 View Post
    Thanks. So is there any way to find the center (r,r) if the only information given is the tangent line 3x+4y-12=0?

    Do you mean a tangent circle to the two axis and also to that equation? If so input the values in the equation for the distance of the point (r,r) to

    the line 3x+4y-12=0:

    \frac{|3r+4r-12|}{\sqrt{3^2+4^2}}=r\iff 49r^2-168r+144= 25r^2\iff r^2-7r+6=0\Longrightarrow r=1,\,6 , so there are two possibilities:

    (x-1)^2+(y-1)^2=1\,,\,\,\,(x-6)^2+(y-6)^2=36

    Tonio

    Ps. There's a mistake above: r=6 is impossible (why?)
    Last edited by tonio; June 19th 2010 at 04:37 AM. Reason: Mistake: r=6 is impossible.
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    Quote Originally Posted by juankim111222 View Post
    Please help me with the process of finding the equation of a circle in the first quadrant that is tangent to the x-axis, y-axis, and a line of the form ax+by+c=0. Thanks.
    if the circle is only in the first quadrant, here is another way to do it.

    See the attachment...
    the triangle contains pairs of identical right-angled triangles,
    hence...

    a-r+b-r=c where c is the length of the hypotenuse, the 3rd side of the triangle.

    a+b-2r=c

    2r=a+b-c

    r=\frac{a+b-c}{2}

    c=\sqrt{a^2+b^2}=\sqrt{3^2+4^2}=5

    Then it is straightforward to find r.

    Tonio's method finds both solutions.
    the circle of r=6 is also tangent to all 3 lines.
    Attached Thumbnails Attached Thumbnails Equation of Circle Tangent to Y-Axis, X-Axis and Line-incircle-radius.jpg  
    Last edited by Archie Meade; June 9th 2010 at 09:47 AM.
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    Re: Equation of Circle Tangent to Y-Axis, X-Axis and Line

    What about the circle to the right of the line, that is also in the first quadrant and tangent to the line and both axes? how do we find its radius?
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    Re: Equation of Circle Tangent to Y-Axis, X-Axis and Line

    Quote Originally Posted by tonio View Post
    ...

    Tonio

    Ps. There's a mistake above: r=6 is impossible (why?)<-- no
    Quote Originally Posted by Archie Meade View Post
    ...

    Tonio's method finds both solutions.
    the circle of r=6 is also tangent to all 3 lines.
    Quote Originally Posted by yvonne1981 View Post
    What about the circle to the right of the line, that is also in the first quadrant and tangent to the line and both axes? how do we find its radius? <--- see Tonio's post
    In addition to the previous posts I've done a geometrical construction: Yo'll find the two centers as Points of intersection of two of the three angle bisectors.

    The colour of the bisector correspond to the colour of the circle. All three angle bisectors form a right triangle.
    Attached Thumbnails Attached Thumbnails Equation of Circle Tangent to Y-Axis, X-Axis and Line-krs3tangenten.png  
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