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 April 4th, 2013, 03:13 AM #1 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 Circle problems 1 a circle is tangent to the two axes and the line 4x+3y-12=0. Find it's equation. 2 consider the circle x^2 + y^2 - 277 = 0. If M(3,-5) is the midpoint of a chord of the circle, find the equation of the chord.
April 4th, 2013, 06:08 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Circle problems

Hello, justusphung!

Did you make any sketches?

Quote:
 $\text{1. A circle is tangent to the two axes and the line }4x\,+\,3y\,-\,12\:=\:0. \;\;\;\text{Find its equation.}$

Use the scroll bar to see the entire diagram.

Code:
      |
4 o
| o
|   o
|     o
|       o
|         o
|           o
|       * * * o
|   *           o
| *               o
|*              /  *o
|             /r      o
*         C /       *   o
* - - - - o         *     o
*    r    :         *       o
|         :                   o
|*        :r       *            o
| *       :       *               o
|   *     :     *                   o
---+-------*-*-*-------------------------o----
|                                     3

$\text{The center is at }C(r,\,r)
\;\;\;\text{where }r\text{ is the radius of the inscribed circle.}
\text{The triangle is a 3-4-5 right triangle.}$

There is a sneaky solution to this problem.

$\text{Formula: }\;\text{Area} \:=\:\frac{1}{2}\text{(perimeter)}\,\times\,r$

$\text{The area is: }\,\frac{1}{2}(3)(4) \:=\:6$

$\text{The perimeter is: }\,3\,+\,4\,+\,5 \:=\:12$

$\text{W\!e have: }\:6 \:=\:\frac{1}{2}(12)r \;\;\;\Rightarrow\;\;\;r \,=\,1$

$\text{The circle is: }\:(x\,-\,1)^2\,+\,(y\,-\,1)^2 \:=\:1^2$

April 4th, 2013, 06:33 AM   #3
Math Team

Joined: Nov 2010
From: Greece, Thessaloniki

Posts: 1,990
Thanks: 133

Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
Re: Circle problems

[color=#000000]Soroban finished it first, but since I made a figure I will upload my solution.

1. Draw the line 4x+3y-12=0, you will notice the right angle triangle with its two perpendicular sides lying on the x and y axis like seen in the next figure.

[attachment=0:1j41gfsh]justusphung.png[/attachment:1j41gfsh]

Since the x-axis, y-axis and the hypoteneuse of the triangle ABC are tangent to the circle, we deduce that the circle we seek is the circle inscibed in the triangle ABC.

So the center of the circle is $\frac{B\cdot |AC|+C\cdot |AB|+A\cdot |AC|}{\text{Perimeter of ABC}}=\frac{(0,4)\cdot 3+(3,0)\cdot 4+(0,0)\cdot 5}{12}=\frac{1}{12}(12,12)=(1,1)$ and since the distance of (1,1) from the x and y axis is 1 the radius we seek is 1. So the circle is $(x-1)^2+(y-1)^2=1$

2. We are given the circle $x^2+y^2=277$, which has center the origin of the axis (0,0). The radius which meets the chord at point M perpendicularly has equation of the form $y=\lambda_{1}x$ since it passes from the origin (0,0). In order to find $\lambda_1$ we plug in the coordinates of the point M, so $\lambda_{1}=-\frac{5}{3}$ and the equation is $y=-\frac{5}{3}x$. Now this line must be perpendicular to the line which is defined by the chord at point M. The line we seek has equation $y-y_0=\lambda_{2}(x-x_0)$, we know that $\lambda_1 \cdot \lambda_2=-1\Rightarrow \lambda_2=\frac{3}{5}$ and for $(x_0,y_0)=(3,-5)$, the equation we seek is $y+5=\frac{3}{5}(x-3)$.[/color]
Attached Images
 justusphung.png (28.5 KB, 213 views)

 April 4th, 2013, 07:15 AM #4 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 Re: Circle problems Thank you for your reply and the precise drawings. By the way, despite of the answer you provided in question 1, there are another 3 possible answers. Can you guys help me find out the other 3 outcomes. Thanks a lot. For question 2, ZardoZ has solved it out. reli thx.
 April 4th, 2013, 07:46 AM #5 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 Re: Circle problems Arr! Finally, I figured it out. As there is one possible answer in section 2 and section 4 of the coordinate plane respectively, while in section 1 there are 2 possible answers, we can derive the equation by determining the distance of the circle centre to the given line, 4x+3y-12=0.
 April 4th, 2013, 07:56 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Circle problems [color=#000000]One other possible way to do it is to use your drawing skills, take a squared paper draw the x and y axis, then the given line and then the bisectors of all the angles of the triangle that soroban and I have mentioned in our solutions. The point where the angle bisectors meet is the centre of the circle you seek.[/color]

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