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Miscellaneous Exercise 6 (II) · Q60

Q.Show that the circles x2+y2+4x−12y+4=0x^2+y^2+4x-12y+4=0 and x2+y2−2x−4y+4=0x^2+y^2-2x-4y+4=0 touch each other internally. Find their point of contact and the equation of their common tangent.

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Circle 1: x2+y2+4x−12y+4=0x^2+y^2+4x-12y+4=0, so g1=2,f1=−6,c1=4g_1=2,f_1=-6,c_1=4; centre C1=(−2,6)C_1=(-2,6), r1=4+36−4=36=6r_1=\sqrt{4+36-4}=\sqrt{36}=6. Circle 2: x2+y2−2x−4y+4=0x^2+y^2-2x-4y+4=0, so g2=−1,f2=−2,c2=4g_2=-1,f_2=-2,c_2=4; centre C2=(1,2)C_2=(1,2), r2=1+4−4=1r_2=\sqrt{1+4-4}=1. Distance C1C2=(−2−1)2+(6−2)2=9+16=5=∣r1−r2∣=∣6−1∣C_1C_2=\sqrt{(-2-1)^2+(6-2)^2}=\sqrt{9+16}=5=|r_1-r_2|=|6-1|. So the circles touch internally.

Point of contact divides C1C2C_1C_2 externally in ratio r1:r2=6:1r_1:r_2=6:1: (6(1)−1(−2)6−1, 6(2)−1(6)6−1)=(6+25,12−65)=(85,65)\left(\dfrac{6(1)-1(-2)}{6-1},\ \dfrac{6(2)-1(6)}{6-1}\right)=\left(\dfrac{6+2}5,\dfrac{12-6}5\right)=\left(\dfrac85,\dfrac65\right). …

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