circlesgeometry
Two circles having radii $7$cm & $19$ cm are separated by a distance of $22$ cm between their centers. If they are intersecting each other at two points $P$ & $Q$ then what will be the length of the common chord PQ?
a.) $\frac{9\sqrt{205}}{13}$
b.) $\frac{8\sqrt{255}}{11}$
c.) $\frac{9\sqrt{233}}{13}$
d.) $\frac{8\sqrt{155}}{11}$
I tried this by assuming one circle to be centered at the origin but it's creating utter mess. I am 12th grade. Thanks for your help!
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.
Don't know that it's much simpler than calculating the pairwise intersections, then the distances to the third center, but the following gives a symmetric condition using complex numbers.
Let $\,a,b,c\,$ be the complex numbers associated with points $A,B,C$ in a complex plane centered at the centroid of $ABC\,$, so that $a+b+c=0\,$.
The point of intersection $z$ of the three circles (if it exists) must satisfy the $3$ equations similar to:
$$ |z-a|^2=R_A^2 \;\;\iff\;\;(z-a)(\bar z - \bar a) = R_A^2 \;\;\iff\;\;|z|^2 - z \bar a - \bar z a + |a|^2 = R_A^2 \tag{1} $$
Writing $(1)$ for $a,b,c$ and summing the $3$ equations up:
$$ \require{cancel} 3\,|z|^2 - \cancel{z \sum_{cyc} \bar a} - \bcancel{\bar z \sum_{cyc} a} + \sum_{cyc}|a|^2 = \sum_{cyc} R_A^2 \;\;\implies\;\; |z|^2 = \frac{1}{3}\left(\sum_{cyc} R_A^2-\sum_{cyc}|a|^2\right) =R^2 \tag{2} $$
Substituting $(2)$ back into each of $(1)\,$:
$$ -|z|^2 + z \bar a + \bar z a - |a|^2 = - R_A^2 \;\;\iff\;\; z \cdot \bar a + \bar z \cdot a = |a|^2+R^2-R_A^2 \tag{3} $$
Considering $(3)$ as a system of linear equations in $z, \bar z\,$, the condition for it to have solutions is:
$$ \left| \begin{matrix} \;\bar a \;&\; a \;&\; |a|^2+R^2-R_A^2\; \\ \;\bar b \;&\; b \;&\; |b|^2+R^2-R_B^2\; \\ \;\bar c \;&\; c \;&\; |c|^2+R^2-R_C^2\; \end{matrix} \right| \;\;=\;\; 0 $$
Best Answer
$$7^2-x^2=19^2-(22-x)^2$$ $$x=\frac{43}{11}$$ so $$h^2=7^2-\frac{43^2}{11^2}=\frac{4080}{121}$$ $$h=\frac{4\sqrt{255}}{11}$$ hence the Chord $PQ$ = $2h$
$$PQ=\frac{8\sqrt{255}}{11}$$