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Getty's question at Yahoo! Answers regarding the distance from a circle where two tangent lines meet
- Thread starter MarkFL
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Hello Getty,
We can greatly simplify this problem, if we orient the circle's center at the origin of our coordinate system, and rotate the circle such that the two tangent points have the same $x$-coordinate (where $0<x<r$), one point in the first quadrant, and one in the fourth quadrant.
Please refer to the following diagram:
Because $r$ and $\ell$ are perpendicular, we may state:
\(\displaystyle r^2+\ell^2=d^2\)
Using the distance formula, we find:
\(\displaystyle \ell^2=(x-d)^2+y^2\)
and from the equation of the circle, we have:
\(\displaystyle y^2=r^2-x^2\)
Hence, we may now write:
\(\displaystyle r^2+(x-d)^2+r^2-x^2=d^2\)
\(\displaystyle 2r^2+x^2-2xd+d^2-x^2=d^2\)
\(\displaystyle r^2-xd=0\)
\(\displaystyle d=\frac{r^2}{x}\)
We can greatly simplify this problem, if we orient the circle's center at the origin of our coordinate system, and rotate the circle such that the two tangent points have the same $x$-coordinate (where $0<x<r$), one point in the first quadrant, and one in the fourth quadrant.
Please refer to the following diagram:
Because $r$ and $\ell$ are perpendicular, we may state:
\(\displaystyle r^2+\ell^2=d^2\)
Using the distance formula, we find:
\(\displaystyle \ell^2=(x-d)^2+y^2\)
and from the equation of the circle, we have:
\(\displaystyle y^2=r^2-x^2\)
Hence, we may now write:
\(\displaystyle r^2+(x-d)^2+r^2-x^2=d^2\)
\(\displaystyle 2r^2+x^2-2xd+d^2-x^2=d^2\)
\(\displaystyle r^2-xd=0\)
\(\displaystyle d=\frac{r^2}{x}\)