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Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show some short-cut to do it quickly

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Let $P$ be the point where the two chords (and a diameter) meet. Let $h$ (and $k$) be the distance from $P$ to the midpoint of the $2a$ chord (respectively, the $2b$ chord); that is, say $P$ divides the chord into sub-segments of length $a+h$ and $a-h$ (respectively, $b+k$ and $b-k$). Note that $P$ divides a diameter into sub-segments of length $r+c$ and $r-c$ (where $r$ is the radius of the circle); note also that $c$ is the hypotenuse of a right triangle with legs $h$ and $k$: so, $c^2 = h^2 + k^2$.

The Power of a Point principle says that every chord through a particular point of a circle is divided into sub-segments such that the product of the lengths of those sub-segments is a constant (the so-called "power" of the point in question). Thus,

$$(a+h)(a-h) = (b+k)(b-k) = (r+c)(r-c)$$

More succinctly,

$$a^2 - h^2 \;\;=\;\; b^2 - k^2 \;\;=\;\; r^2 - c^2$$

With an eye towards combining an $h^2$ with a $k^2$, I'll add the left-hand and "middle-hand" sides together; their sum is necessarily twice the right-hand side:

$$\begin{align} ( a^2 - h^2 ) + ( b^2 - k^2 ) &= 2 (r^2 - c^2) \\ \implies a^2 + b^2 - ( h^2 + k^2 ) &= 2 r^2 - 2 c^2 \\ \implies a^2 + b^2 - c^2 &= 2 r^2 - 2 c^2 \\ \implies a^2 + b^2 + c^2 &= 2 r^2 \end{align}$$

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Another similar approach may be arisen as following: ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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  • $\begingroup$ I got it but can't I solve without incorporating axis system? $\endgroup$ – iostream007 May 17 '13 at 7:37
  • $\begingroup$ @iostream007: I did it without Calculus. I made $xy$ axes just to make the solution better. You can remove it easily. Just know that two chords are perpendicular to diameters of the circle and so the diameter bisect them to $a=AH$ and $b=DL$ respectively. $\endgroup$ – Mikasa May 17 '13 at 7:42
  • $\begingroup$ @iostream007: $a^2+b^2+c^2=2r^2$ $\endgroup$ – Mikasa May 17 '13 at 7:44
  • $\begingroup$ nice work, dear friend! $\;\;+\;\left(\lim_{n\to\infty} \sqrt[\Large n]{n}\right)\;\;$ $\endgroup$ – amWhy May 18 '13 at 1:25
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Let $a'$, $b'$ be the lengths of the perpendiculars from $M$ to the chord of length $2a$, resp. $2b$. Then $$c^2=a'^2+b'^2=(r^2-a^2)+(r^2-b^2)\ ,$$ from which we obtain $$r=\sqrt{a^2+b^2+c^2\over2}\ .$$

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Draw a circle through c. Let's suppose the chords are parallel to the x and y axis, and that the points a and b lie on some large circle, and the lines $x=a$ and $y=b$ cross at $A, B$

We then have the equation that $A^2+B^2=C^2$ for the crossing, which gives these lines, and thence $A^2 + b^2 = a^2 + B^2 = R^2$, which leads to the equation $a^2+b^2+c^2 = 2R^2$.

Answer: $R^2 = (a^2+b^2+c^2)/2$

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  • $\begingroup$ How t0 do it for general case? $\endgroup$ – iostream007 May 17 '13 at 7:27
  • $\begingroup$ @isostream007. It is a general case. One should note that we simply set the coordinates up to simplify the coordinates. The coordinates where the chords strike the sphere is at $A, b$ and $a,B$, and the lines cross at $A,B$ at distance $c$. Because we know that $A^2+b^2 = a^2 + B^2 = R^2$, and $A^2+B^2=C^2$, the sum of the first two LHS, with the third substituted of the second reduces the value of R directly in terms of a, b, c. $\endgroup$ – wendy.krieger May 17 '13 at 8:20
  • $\begingroup$ The relationship of this equation is identical to that of a a tetrahedron, whose opposite sides are equal. $\endgroup$ – wendy.krieger May 17 '13 at 8:23

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