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"Two circles with centres C1 and C2 and radius 6 cm and 8 cm respetively cut each other at right angles. Find the length of the common chord."

I tried it but could not get to the answer.

I am not getting that circles cut each other at right angles.

So do we need to make a tangent that will intersect at right angle?

Any suggestions welcome. Thanks in advance.

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3 Answers 3

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Draw the two circles; since they cut at right angles, the radius on $C_1$ that goes through a point of intersection meets the radius of $C_2$ that goes through that point of intersection at a right angle. This gives you a right triangle formed by the two radii and the line joining the centers of $C_1$ and $C_2$, which must therefore by of length $\sqrt{6^2+8^2} = \sqrt{100}=10$.

The common chord then has twice the length of the height of a 6-8-10 right triangle that is resting on its hypothenuse. Since such a triangle has area $\frac{1}{2}(6)(8) = 24$, the height when it is resting on its hypothenuse is $\frac{24}{5}$; so the common chord has length $\frac{48}{5}$ cm.

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2  
I was going to answer similarly. Here is a diagram. –  robjohn Mar 31 '12 at 4:25
    
@Arturo thanks.got it.:) –  vikiiii Mar 31 '12 at 13:13

Let $A$ and $B$ be the two points of intersection. The angles $C_1AC_2$ and $C_1BC_2$ are right, hence the segments $C_1A,\quad$ $C_2A, \quad$ $C_1B \quad$ and $C_2B$ are the radii of their respective circles. This means that their lengths are $6$ and $8$ respectively. The triangle $C_1A C_2 C_1$ is right at $A$, hence by Pythagoras' theorem the length $C_1 C_2$ is $10$. Now the segment $AB$ has to be perpendicular to the segment $C_1C_2$, so that the area of the triangle $C_1 A C_2 C_1$ can be computed in two ways : $\frac 12 \cdot 6 \cdot 8$ or $\frac 12 \cdot 10 \cdot (\frac 12 \ell)$, where $\ell$ is the length of the common chord. Computing gives you $$ \frac 12 \cdot 6 \cdot 8 = \frac 12 \cdot 10 \cdot \frac 12 \ell, \quad \Longrightarrow \ell = 9.6 $$ Hope that helps,

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Find out the radical axis of the two circles. This is nothing but the common chord of the two circles. Substitute for $x$ or $y$ in any of the equation of the circle from the radical axis. You will get two points of intersection, and then you can apply the distance formula you will get the length of the common chord.

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Try not to skip letters when forming words. Coherent sentences and proper grammar are usually expected here. –  Joe Apr 1 '12 at 20:10
    
I undertook the task of translating that mess into (mostly) standard English. Joe's comment refers to the previous version. –  rschwieb Aug 17 '12 at 13:19

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