direct common tangent

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by hemangsarkar, Sep 6, 2012, 6:18 AM

i read the result related to this somewhere. and thought of proving it.

Q) find the length of portion of the direct common tangent of two given circles between them. given that $r_{1}$ and $r_{2}$ are their radii and $d$ is the distance between their centers.

solution -

consider two circles. one of them centered at $O$ , $(0,0)$ with radius $r_{1}$ and the other centered at point $M$ , $(d,0)$ with radius $r_{2}$ .

let $l$ be the direct common tangent. it touches them at $A$ and $B$ respectively.
let $l$ intersect the $x$ axis at $P$ .

it can be shown that -
$P = \left( \frac{r_{1}d}{r_{1} - r_{2}} , 0 \right)$

using pythagoras theorem in the triangles, we get
$AP = \sqrt{\left(\frac{r_{1}d}{r_{1}-r_{2}} \right)^2 - r_{1}^2}$

and $BP = \sqrt{\left(\frac{r_{1}d}{r_{1}-r_{2}} - d \right)^2 - r_{2}^2}$

so, $AP - BP = AB = \sqrt{d^2 - (r_{1} - r_{2})^2}$

this is the length of the portion of the direct common tangent between the circles.

if $r_{1} = r_{2}$ , then $AB = d$ .

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direct common tangent

by hemangsarkar, Sep 6, 2012, 6:18 AM

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