circlesgeometry
"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.
Let us denote the radius of circle with center $A$ as $r_{1}$ and circle with center $C$ as $r_{2}$, and $AC=d$ as the distance between the centers.
Now note by SAS criteria (prove $\triangle ADC$ and $\triangle ABC$ are congruent by SSS and ...) that $\triangle ABE$ and $\triangle ADE$ are congruent. Thus $\angle AEB = \angle AED$, but since $\angle AEB + \angle AED = 180^{\circ}$, we have $\angle AEB = \angle AED = 90^{\circ}$. Similarly, $\angle BEC = \angle DEC= 90^{\circ}$.
Thus, we can see that $BE$ is the height of $\triangle ABC$, and $DE$ is the height of $\triangle ADC$, with base $AC$. So,
$$BD=BE+DE=\frac{2(ABC)}{AC} + \frac{2(ADC)}{AC}=\frac{4}{d}(ABC)$$
Where $(ABC)$ and $(ADC)$ represents the area of the respective triangles, which we can calculate in terms of $r_{1},r_{2},d$ using the Herons formula. This is the general formula which gives the length of the common chord, $BD$. I leave the proof of the above formula above as a exercise for you.
However, your formula was for a special case, when $\angle ABC = \angle ADC= 90^{\circ}$. This simplifies the area of the triangles and gives:
$$BC=\frac{4}{d}(ABC)=\frac{2r_{1}r_{2}}{d}$$
Since, in your first question, the pair $(r_{1},r_{2},d)$ was $(15,20,25)$, which is a Pythagorean triple, so the condition $\angle ABC = \angle ADC= 90^{\circ}$ held, and so did your formula. Now lets come to the second question, with a new diagram:
I leave it as a exercise to you to prove that $r_{1}=r_{2}=AB=AC=BC$. Hence, $ABC$ is an equilateral triangle with $\angle ABC = 60^{\circ} \neq 90^{\circ}$. Hence your formula does not hold, and we must return to the general formula:
$$AD=\frac{4}{d}(ABC)=\frac{4}{d}\frac{\sqrt{3}(AC)^2}{4}=\frac{\sqrt{3}(r_{1})^2}{r_{1}}=\sqrt{3}r_{1}$$
What I essentially did in the second last step, was to use the Pythagoras theorem, like the way you must have did, to find the altitude in terms of the side, $r_{1}$, and use it to find the area of the triangle, or we can directly find $AD$ from the height.
$$7^2-x^2=19^2-(22-x)^2$$ $$x=\frac{43}{11}$$ so $$h^2=7^2-\frac{43^2}{11^2}=\frac{4080}{121}$$ $$h=\frac{4\sqrt{255}}{11}$$ hence the Chord $PQ$ = $2h$
$$PQ=\frac{8\sqrt{255}}{11}$$
Best Answer
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,