Circle circled circles

Let $r_i$ be the radius of circle $\Gamma_i$ . We have \ $r_1 = r_2 + r_3$ , so the area that we are interested in is $\pi r_1 ^2 - pi r_2 ^2 - \pi r_3 ^2 = \pi 2 r_2 r_3$ . By similar triangles of $ACP$ and $PCB$ , we have $2r_2 \times 2 r_3 AC \times CB = PC \times PC = 10^2$ . Hence, $M = \frac {1}{2} ( 2 r_2 \times 2 r_3) = 50$ .

First, notice that triangle APC and triangle PBC is similar because $\angle ACP=\angle BCP=90^\circ$ and $\angle APC=180^\circ-90^\circ-\angle PAC=90^\circ-\angle APC=\angle PBA$ . So, $(AC)(BC)=100$ . $(BC)=100/(AC)$ . The area of $\Gamma1$ is $\pi r^2$ = $\pi {\frac{(AC)^2+100}{2(AC)}}^2$ = $\pi\frac{(AC)^4+10000+200(AC)^2}{4(AC)^2}$ The area of $\Gamma2$ = $\pi{\frac{(AC)}{2}}^2$ = $\pi\frac{(AC)^2}{4}$ = $\pi\frac{(AC)^4}{4(AC)^2}$ The area of $\Gamma3$ = $\pi{\frac{(BC)}{2}}^2$ = $\pi[{\frac{100}{2(AC)}}]^2$ = $\pi{\frac{10000}{4(AC)^2}}$ M $\pi$ =area of $\Gamma1$ -area of $\Gamma2$ -area of $\Gamma3$ = $\pi\frac{(AC)^4+10000+200(AC)^2}{4(AC)^2}$ - $\pi\frac{(AC)^4}{4(AC)^2}$ - $\pi{\frac{10000}{4(AC)^2}}$ = $\pi\frac{200(AC)^2}{4(AC)^2}$ = $50\pi$ Therefore, M=5.

Let the center of the large circle be $O$ , let the center of $Γ_2$ be $M$ , and let the center of $Γ_3$ be $N$ . Since $Γ_1$ , $Γ_2$ , and $Γ_3$ are pairwise tangent, we get that $A$ , $M$ , $N$ , and $B$ are co-linear.

Let $PQ$ intersect $AB$ at point $K$ . Then, $PQ$ is perpendicular to $AB$ [since $PQ$ is tangent to both $Γ_2$ and $Γ_3$ ], which implies $K$ is the midpoint of $PQ$ . Now let the radius of the largest circle be $r_1$ , let the radius of $Γ_2$ be $r_2$ , and let the radius of $Γ_3$ be $r_3$ .

Note that $AB= AK + KB \implies 2r_1 = 2r_2 + 2r_3 \implies r_1 = r_2 + r_3$ . Now applying secant rule on the intersecting chords $PQ$ and $AB$ , we get $AK. BK = PK.KQ \implies 2r_2 * 2r_3 = 10^2 \implies 4r_2(r_1 - r_2) = 100 \implies r_2r_1 - r_2^2 = 25$ .

Now the required area= $area of Γ_1 - area of Γ_2 - area of Γ_3 = \pi (r_1^2 - r_2^2 - (r_1 - r_2)^2)$ $= \pi (r_1^2 - r_2^2 - r_1^2 - r_2^2 + 2r_1r_2) = \pi( 2(r_1r_2 - r_2^2 ))$ $= \pi (2*25) = 50 \pi$ . Thus $M= 50$ .

Suppose $C$ does not lie on $AB$ . Then, the reflection of $C$ across the line $AB$ would be another point where $\Gamma_2, \Gamma_3$ are tangential. Since 2 distinct circles cannot be tangent at 2 points, we must have that points $A, B$ and $C$ are collinear.

Let $O$ be the center of $\Gamma_1$ . Let $\Gamma_2$ and $\Gamma_3$ have radius $R$ and $r$ , respectively. Without loss of generality, let $R \geq r$ . From the above observation, the radius of $\Gamma_1$ is $R+r$ . The area that we’re interested in has the form $\pi (R+r)^2 - \pi R^2 - \pi r^2 = 2Rr\pi$ , thus $M = 2Rr$ .

Consider triangle $OCP$ . It is right angled with $OC = R + r - 2r = R - r,$ $OP = R + r$ and $CP = \frac{20}{2} = 10$ . Applying the Pythagorean theorem, we get that $(R+r)^2 = (R-r)^2 + 10 ^2$ . Expanding gives $4Rr = 10 ^2 \Rightarrow Rr=25$ . Hence $M =2Rr = 50$ .

Let c is the centre of the big circle.Then required area =22/7 ×(10)^2- 2×22/7 ×(5)^2=50 (22/7).then M=50.

Drawing the full figure, being R1, R2 and R3 the radio from circunferences 1, 2 and 3, respectively, and drawing the triangle PAB (or QAB), we are going to obtain a equation system to solve M.

• (1) First, we use trigonometry. Watching PAB is simple that $\tan \angle PA = \frac{\frac{PQ}{2}}{2 * R2}$ and $\tan \frac{\pi}{2} - \angle PA = \frac{\frac{PQ}{2}}{2 * R3}$

This is: $\tan \angle PA = \frac{10}{2*R2}$ and $\tan \frac{\pi}{2} - \angle PA= \frac{10}{2 * R3}$ Following trigonometry reason: $\tan A = \frac{1}{\tan \frac{\pi}{2} - A}$

we solve:

$R2 * R3 = 25$ (simplifying)

• (2) In other way, the area asked is

$\pi * (R1^2 - R2^2 - R3^2) = M * \pi$

• (3) From the direct relationships between radios (watching full figure): $2 * R1 = 2 * (R2 + R3)$ , we substitute and simplify using equation from (2):

$2 * R2 * R3 = M$

We substitute the previous equation obtained from (1), trigonometry, $R2 * R3 = 25)$ , so: $2 * 25 = M$

$M=50$

Since circle $Γ_2$ and circle $Γ_3$ are internally tangent to the circle $Γ_1$ at the end points of the diameter (A,B) .and they are externally tangent at point C,therefore $r_1=r_2+r_3$ * (1) * . The area needed = $pi*(r_1^2-r_2^2-r_3^2)$ , therefore M= $r_1^2-r_2^2-r_3^2$ . squaring both sides of equation (1) we get $r_1^2-r_2^2-r_3^2=2*r_1*r_2$ . $PQ=20$ due to symmetry $PC$ =10 , referring to triangle PAB $\angle APB = 90 ^ \circ$ and $\angle PCA = 90 ^ \circ$ as the line PC is a tangent therefore $PC^2=AC * CB$ , $4*r_1*r_2=100$ , $2*r_1*r_2=50$ , $M=2*r_1*r_2=50$ . This can be proved using Pythagorean theorem Assume we have a right-angled triangle ABC at A and assume we have the height drawn to cut the hypotenuse has length z and it cuts it at D such that BD= x DC= y and let AB= c AC= b we have $c^2+b^2=(y+x)^2$ , $c^2+b^2=y^2+2yx+x^2$ (1) In triangle ADB $x^2+z^2=c^2$ (2) ,In triangle ADC $b^2=y^2+z^2$ (3) , adding (2),(3) we get $c^2+b^2=x^2+y^2+2z^2$ (4) subtracting (4) from (1) $xy=z^2$

Let the radius of larger circle be a, smaller ones be 'b' and 'c'. Ans = pie(a^2-(b^2+c^2)) here we are only concerned with (a^2-(b^2+c^2))= (a^2 - ((b + c)^2 - 2bc) = 2bc as a=b+c Using intersecting chord theorem, 2bc= AC BC/4= PC CQ/4= 10*10/4 as diameter bisects chord = 50 Ans

Let's denote the radius of the circles $\Gamma_{2}$ and $\Gamma_{3}$ are $a$ and $b$ respectively. Hence, the radius of $\Gamma_{1}$ is $a+b$ ; and $M\pi=\pi((a+b)^2-a^2-b^2)=2\pi ab\Rightarrow M=2ab$ . On the other hand, APB is a right triangle with altitude $PC$ , so $PC^2=AC*BC$ i.e $(\frac{PQ}{2})^2=2a*2b \Rightarrow ab=\frac{PQ^2}{16}=25$ . So $M=2ab=50$

Let the radius of larger circle be a, smaller ones be 'b' and 'c'. Ans = pie(a^2-(b^2+c^2)) here we are only concerned with (a^2-(b^2+c^2))= (a^2 - ((b + c)^2 - 2bc) = 2bc as a=b+c Using intersecting chord theorem, 2bc= AC BC/4= PC CQ/4= 10*10/4 as diameter bisects chord = 50 Ans

a = radius of gamma1 b = radius of gamma2 c = radius of gamma3 a = b+c

Area1 = (b^2 + 2bc + c^2) * pi Area2 = (b^2) * pi Area3 = (c^2) * pi AreaWeWant = Area1 - Area2 - Area3 = (2bc) * pi

If you draw it out, you will see that we have two chords AB and PQ AB is composed of segments AC and CB with lengths 2b and 2c PQ is composed of segments PC and CQ with lengths 10 and 10 For chords in circles, (2b) (2c) = (10) (10) 4bc = 100 bc = 25

AreaWeWant = (2bc) * pi = 50 pi M = 50

Let the radius of $Γ_2$ ­ be r and that of $Γ_3$ be (R-r). Let the centre of $Γ_1$ be O. PQ=20, hence, $\frac {1}{2}$ PQ=10. Let PQ meet AB at K. Therefore, from $\triangle OKP$ we get: $OK^2$ + $KP^2$ = $OP^2$ Hence, $(2r-R)^2$ + $10^2$ = $R^2$ ; or, r(R-r)=25............(1)

The area that is within $Γ_1$ but not in $Γ_2$ or $Γ_3$ is equal to $\pi$ [ $R^2$ - $(R-r)^2$ - $r^2$ ] = 2 $\pi$ r(R-r) = 2 $\pi \times 25$ = 50 $\pi$ [From (1)]

Hence, M=50.

Let R be the radius Γ2 and r the radius of Γ3.

The radius of Γ1 is (2R+2r)/2=R+r.

The required area is then:

π (R+r)^2-π r^2-π R^2 = 2πrR.

The triangle ABQ is right and for the second theoreme of Euclide:

2r*2R=QC^2.

So:

4rR=QC^2=(QP/2)^2=100.

So 2πrR = 50π.

The required area is 50π, and so M=50.