Prove that the circle $x^{2} + y^{2} - 9 = 4 r^{2}, r = 1, 2, 3$ cut off equal interception between the circles on the line $3 x + 4 y + 15 = 0$

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Solution

$\begin{matrix} x^{2} + y^{2} = 9 + 4 r^{2} Δ A O M ∠ O M A = 90^{0} O A^{2} = O B^{2} + A M^{2} O M = \frac{3 (0) + 4 (0) + 15}{\sqrt{3^{2} + 4^{2}}} = \frac{15}{5} = 3 \sqrt{9 + 4 r^{2}} = 3^{2} + A M^{2} 9 + 4 r^{2} = 9 + A M^{2} A M = 2 r A M = M B = 2 (1) = 2 A B = 4 M C = M D (r = 2) C D = 8 M E = M F M F = 12 C A = M C - M A = 4 - 2 = 2 B O = M D - M B = 4 - 2 = 2 M C = M E - M C = 6 - 4 = 2 D E = M F - M D = 6 - 4 = 2 A C = B D = E C = D F \end{matrix}$

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Prove that the circle x2+y2−9=4r2,r=1,2,3 cut off equal interception between the circles on the line 3x+4y+15=0

x2+y2=9+4r2ΔAOM∠OMA=900OA2=OB2+AM2OM=3(0)+4(0)+15√32+42=155=3√9+4r2=32+AM29+4r2=9+AM2AM=2rAM=MB=2(1)=2AB=4MC=MD(r=2)CD=8ME=MFMF=12CA=MC−MA=4−2=2BO=MD−MB=4−2=2MC=ME−MC=6−4=2DE=MF−MD=6−4=2AC=BD=EC=DF

Prove that the circle $x^{2} + y^{2} - 9 = 4 r^{2}, r = 1, 2, 3$ cut off equal interception between the circles on the line $3 x + 4 y + 15 = 0$