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The Art and Craft of Problem Solving, 2nd Edition

By: Paul Zeitz

1.3.4

In the xy-plane, what is the length of the shortest path from [tex](0,0)[/tex] to [tex](12, 16)[/tex] that does not go inside the circle [itex]{{(x - 6)}^{2}} + {{(y - 8)}^{2}} = {25}[/itex].

Standard form of the Equation of a Circle.

[tex]

{{(x - h)}^{2}} + {{(y - k)}^{2}} = {{r}^{2}}

[/tex]

Standard form of the distance between two points [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex] in the xy-plane.

[tex]

{d} = {\sqrt {{{{\left({x}_{2}} - {{x}_{1}}\right)}^{2}} + {{\left({{y}_{2}} - {{y}_{1}}\right)}^{2}}}}

[/tex]

[tex]{P}_{m} \equiv[/tex] midpoint

[tex]{d} \equiv[/tex] the distance between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex]

[tex]{d}_{r} \equiv[/tex] diameter

[tex]{{D}_{min}} \equiv[/tex] the length of the shortest path between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex] that does not go inside the circle, [tex]{{(x - 6)}^{2}} + {{(y - 8)}^{2}} = {25}[/tex].

[tex]{C} \equiv[/tex] circumference of the circle.

[tex]({x}_{1}, {y}_{1}) = (0, 0)[/tex]

[tex]({x}_{2}, {y}_{2}) = (12, 16)[/tex]

[tex](h, k) = (6, 8)[/tex]

[tex]{r} = 5[/tex]

Let, [tex]Ax + By = C[/tex] be the equation of the line between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex].

Note, that [tex]h[/tex] and [tex]k[/tex] in this problem are half of [itex]{x}_{2}[/itex] and [itex]{y}_{2}[/itex], additionally since [tex]({x}_{1}, {y}_{1}) = (0, 0)[/tex] the point [tex](h, k)[/tex] is the [tex]{P}_{m}[/tex] of the line [tex]Ax + By = C[/tex]. Furthermore, this means that the length between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex] that goes inside the circle is the diameter. Lastly, this implies that the shortest length to avoid the inside of the circle is half of the circumference of the circle.

So, this problem is then simplified to finding the below,

[tex]

{D}_{min} = {{d} - {{d}_{r}} + {\frac{C}{2}}}

[/tex]

Which is,

[tex]

{{D}_{min}} = {\left({\sqrt{{{\left({{x}_{2}}-{{x}_{1}}\right)}^{2}}+{{\left({{y}_{2}}-{{y}_{1}}\right)}^{2}}}}\right)}-{\left(2r\right)}+{\left(\frac{(2{\pi}{r})}{2}\right)}

[/tex]

[tex]

{{D}_{min}} = {{\sqrt{{{\left({{x}_{2}}-{{x}_{1}}\right)}^{2}}+{{\left({{y}_{2}}-{{y}_{1}}\right)}^{2}}}}}+{r\left({\pi}-2\right)}[/tex]

[tex]

{{D}_{min}} = {10+5{\pi}}

[/tex]

Is the above answer right?

In the xy-plane, what is the length of the shortest path from [tex]({x}_{1}, {y}_{1})[/tex] to [tex]({x}_{2}, {y}_{2})[/tex] that does not go inside the circle [itex]{{(x-h)}^{2}}+{{(y-k)}^{2}} = {{r}^{2}}[/itex].

I tried figuring out the above but did not know how to approach the case if the circle was not "in the way" so to speak of the shortest length between the two points. How is it that one can apply constraints to this problem to tell whether the circle: would be or would not be in the way?

Thanks,

-PFStudent

By: Paul Zeitz

**1. Homework Statement**1.3.4

In the xy-plane, what is the length of the shortest path from [tex](0,0)[/tex] to [tex](12, 16)[/tex] that does not go inside the circle [itex]{{(x - 6)}^{2}} + {{(y - 8)}^{2}} = {25}[/itex].

**2. Homework Equations**Standard form of the Equation of a Circle.

[tex]

{{(x - h)}^{2}} + {{(y - k)}^{2}} = {{r}^{2}}

[/tex]

Standard form of the distance between two points [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex] in the xy-plane.

[tex]

{d} = {\sqrt {{{{\left({x}_{2}} - {{x}_{1}}\right)}^{2}} + {{\left({{y}_{2}} - {{y}_{1}}\right)}^{2}}}}

[/tex]

**3. The Attempt at a Solution**[tex]{P}_{m} \equiv[/tex] midpoint

[tex]{d} \equiv[/tex] the distance between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex]

[tex]{d}_{r} \equiv[/tex] diameter

[tex]{{D}_{min}} \equiv[/tex] the length of the shortest path between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex] that does not go inside the circle, [tex]{{(x - 6)}^{2}} + {{(y - 8)}^{2}} = {25}[/tex].

[tex]{C} \equiv[/tex] circumference of the circle.

[tex]({x}_{1}, {y}_{1}) = (0, 0)[/tex]

[tex]({x}_{2}, {y}_{2}) = (12, 16)[/tex]

[tex](h, k) = (6, 8)[/tex]

[tex]{r} = 5[/tex]

Let, [tex]Ax + By = C[/tex] be the equation of the line between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex].

Note, that [tex]h[/tex] and [tex]k[/tex] in this problem are half of [itex]{x}_{2}[/itex] and [itex]{y}_{2}[/itex], additionally since [tex]({x}_{1}, {y}_{1}) = (0, 0)[/tex] the point [tex](h, k)[/tex] is the [tex]{P}_{m}[/tex] of the line [tex]Ax + By = C[/tex]. Furthermore, this means that the length between [tex]({x}_{1}, {y}_{1})[/tex] and [tex]({x}_{2}, {y}_{2})[/tex] that goes inside the circle is the diameter. Lastly, this implies that the shortest length to avoid the inside of the circle is half of the circumference of the circle.

So, this problem is then simplified to finding the below,

[tex]

{D}_{min} = {{d} - {{d}_{r}} + {\frac{C}{2}}}

[/tex]

Which is,

[tex]

{{D}_{min}} = {\left({\sqrt{{{\left({{x}_{2}}-{{x}_{1}}\right)}^{2}}+{{\left({{y}_{2}}-{{y}_{1}}\right)}^{2}}}}\right)}-{\left(2r\right)}+{\left(\frac{(2{\pi}{r})}{2}\right)}

[/tex]

[tex]

{{D}_{min}} = {{\sqrt{{{\left({{x}_{2}}-{{x}_{1}}\right)}^{2}}+{{\left({{y}_{2}}-{{y}_{1}}\right)}^{2}}}}}+{r\left({\pi}-2\right)}[/tex]

[tex]

{{D}_{min}} = {10+5{\pi}}

[/tex]

Is the above answer right?

**In addition, how would one solve a generalized form of this problem,**In the xy-plane, what is the length of the shortest path from [tex]({x}_{1}, {y}_{1})[/tex] to [tex]({x}_{2}, {y}_{2})[/tex] that does not go inside the circle [itex]{{(x-h)}^{2}}+{{(y-k)}^{2}} = {{r}^{2}}[/itex].

I tried figuring out the above but did not know how to approach the case if the circle was not "in the way" so to speak of the shortest length between the two points. How is it that one can apply constraints to this problem to tell whether the circle: would be or would not be in the way?

Thanks,

-PFStudent

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