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# I am going in circles...

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A circle with center C is tangent to the positive x- and y-axes and is externally tangent to the circle centered at (3,0) with a radius of 1. What is the sum of all possible radii of the circle with center C?

Feb 2, 2019

#1
+2343
+3

I think a diagram would allow you to visualize the problem.

Let r equal the length of the radius of circle C.

The restrictions on circle C are such that the center is equidistant from both axes and lies in the first quadrant of the coordinate plane. I have defined the length of the radius to be r, so the center is units away from both axes; therefore, the coordinates of the center are $$(r,r)$$ .

I am going to represent the distance from both centers in two different ways.

Since circle C and circle A are externally tangent, they intersect at only one common point.  $$l_{\overline{BC}}=r$$ because that segment is another radius, and all radii have the same length. $$l_{\overline{BA}}=1$$  because the radius of circle A is given as 1.

$$l_{\overline{AC}}=l_{\overline{BC}}+l_{\overline{BA}}=r+1$$

$$r+1$$ is the first way to represent the distance between the center of both circles. The second way requires the distance formula.

 $$\text{distance between }(r,r)\text{ and }(3,0);\\ d_{\overline{AC}}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ (3,0) is the coordinate of the center of the circle A. (r,r) is the coordinate of the center of the circle C. $$d_{\overline{AC}}=\sqrt{(r-3)^2+r^2}$$ This is the second way to represent the distance between the center of both circles.

$$r+1\text{ and }\sqrt{(r-3)^2+r^2}$$ represent the same idea: the distance between the center of both circles. This means that they are equal.

 $$\sqrt{(r-3)^2+r^2}=r+1$$ Before beginning to solve this problem, we must compute the domain restriction on this equality since a square root is present. $$(r-3)^2+r^2\geq 0\\ r^2-6r+9+r^2\geq 0\\ 2r^2-6r+9\geq 0$$ Expand this inequality fully on the left side. $$a=2, b=-6, c=9\\ \Delta=b^2-4ac\\ \Delta=(-6)^2-4*2*9\\ \Delta=-36\\ ∴ \text{No restriction in domain}\\ D:(-\infty,\infty)$$ Delta represents the discriminant of this quadratic inequality. This quadratic has a positive leading coefficient, so the corresponding parabola faces upward in the coordinate plane. $$\Delta<0$$, which means that this parabola never intersects the x-axis. The only way for both conditions to be true is that the parabola is forever above the x-axis, so the function will always output a value greater than zero. $$\sqrt{(r-3)^2+r^2}=r+1$$ Solve for r by using algebraic manipulation. $$(r-3)^2+r^2=(r+1)^2\\ (r-3)^2+r^2-(r+1)^2=0\\ r^2-6r+9+r^2-(r^2+2r+1)=0\\ 2r^2-6r+9-r^2-2r-1=0\\ r^2-8r+8=0$$ Square both sides to eliminate the radical, bring all terms to the left-hand side of the equation, and simplify the left-hand side of the equation completely. This quadratic is not factorable, so we will have to resort to other methods to find the solutions. $$a=1, b=-8, c=8\\ r_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ r_{1,2}=\frac{-(-8)\pm\sqrt{(-8)^2-4*1*8}}{2*1}\\ r_{1,2}=\frac{8\pm\sqrt{32}}{2}\\ r_{1,2}=\frac{8\pm4\sqrt{2}}{2}\\ r_{1,2}=4\pm2\sqrt{2}\\ r_1=4+2\sqrt{2}\text{ or }r_2=4-2\sqrt{2}$$ Use the quadratic formula to determine both solutions. Both solutions are positive, so they are both valid for the context of this problem. $$r_1+r_2\\ 4+2\sqrt{2}+4-2\sqrt{2}\\ 8$$ The sum of all possible radii is 8.

Feb 3, 2019

#1
+2343
+3

I think a diagram would allow you to visualize the problem.

Let r equal the length of the radius of circle C.

The restrictions on circle C are such that the center is equidistant from both axes and lies in the first quadrant of the coordinate plane. I have defined the length of the radius to be r, so the center is units away from both axes; therefore, the coordinates of the center are $$(r,r)$$ .

I am going to represent the distance from both centers in two different ways.

Since circle C and circle A are externally tangent, they intersect at only one common point.  $$l_{\overline{BC}}=r$$ because that segment is another radius, and all radii have the same length. $$l_{\overline{BA}}=1$$  because the radius of circle A is given as 1.

$$l_{\overline{AC}}=l_{\overline{BC}}+l_{\overline{BA}}=r+1$$

$$r+1$$ is the first way to represent the distance between the center of both circles. The second way requires the distance formula.

 $$\text{distance between }(r,r)\text{ and }(3,0);\\ d_{\overline{AC}}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ (3,0) is the coordinate of the center of the circle A. (r,r) is the coordinate of the center of the circle C. $$d_{\overline{AC}}=\sqrt{(r-3)^2+r^2}$$ This is the second way to represent the distance between the center of both circles.

$$r+1\text{ and }\sqrt{(r-3)^2+r^2}$$ represent the same idea: the distance between the center of both circles. This means that they are equal.

 $$\sqrt{(r-3)^2+r^2}=r+1$$ Before beginning to solve this problem, we must compute the domain restriction on this equality since a square root is present. $$(r-3)^2+r^2\geq 0\\ r^2-6r+9+r^2\geq 0\\ 2r^2-6r+9\geq 0$$ Expand this inequality fully on the left side. $$a=2, b=-6, c=9\\ \Delta=b^2-4ac\\ \Delta=(-6)^2-4*2*9\\ \Delta=-36\\ ∴ \text{No restriction in domain}\\ D:(-\infty,\infty)$$ Delta represents the discriminant of this quadratic inequality. This quadratic has a positive leading coefficient, so the corresponding parabola faces upward in the coordinate plane. $$\Delta<0$$, which means that this parabola never intersects the x-axis. The only way for both conditions to be true is that the parabola is forever above the x-axis, so the function will always output a value greater than zero. $$\sqrt{(r-3)^2+r^2}=r+1$$ Solve for r by using algebraic manipulation. $$(r-3)^2+r^2=(r+1)^2\\ (r-3)^2+r^2-(r+1)^2=0\\ r^2-6r+9+r^2-(r^2+2r+1)=0\\ 2r^2-6r+9-r^2-2r-1=0\\ r^2-8r+8=0$$ Square both sides to eliminate the radical, bring all terms to the left-hand side of the equation, and simplify the left-hand side of the equation completely. This quadratic is not factorable, so we will have to resort to other methods to find the solutions. $$a=1, b=-8, c=8\\ r_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ r_{1,2}=\frac{-(-8)\pm\sqrt{(-8)^2-4*1*8}}{2*1}\\ r_{1,2}=\frac{8\pm\sqrt{32}}{2}\\ r_{1,2}=\frac{8\pm4\sqrt{2}}{2}\\ r_{1,2}=4\pm2\sqrt{2}\\ r_1=4+2\sqrt{2}\text{ or }r_2=4-2\sqrt{2}$$ Use the quadratic formula to determine both solutions. Both solutions are positive, so they are both valid for the context of this problem. $$r_1+r_2\\ 4+2\sqrt{2}+4-2\sqrt{2}\\ 8$$ The sum of all possible radii is 8.

TheXSquaredFactor Feb 3, 2019
#2
0

I got the r+1 part. I wish I had thought of using the distance formula.

Thank you.

Guest Feb 3, 2019
#3
+103140
0

Very nice, Melody....this is an interesting problem....

CPhill  Feb 3, 2019
#4
+103709
+1

I can not take the credit Chris.

This is x^2 solution. A very nice solution it is too :)

Great work X-squared!

Melody  Feb 3, 2019
#5
+103140
+1

Ah....I missed his avatar on this one....!!!

Yes.....this is very  nice

Just on intuition, it might appear that there are more circles that would qualify....but X'2 shows that only two exist...

CPhill  Feb 3, 2019