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Miscellaneous Exercise 6 (II) · Q58

Q.Show that the circles x2+y2−4x−10y+19=0x^2+y^2-4x-10y+19=0 and x2+y2+2x+8y−23=0x^2+y^2+2x+8y-23=0 touch each other externally. Find their point of contact and the equation of their common tangent.

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Circle 1: x2+y2−4x−10y+19=0x^2+y^2-4x-10y+19=0, so g1=−2,f1=−5,c1=19g_1=-2,f_1=-5,c_1=19; centre C1=(2,5)C_1=(2,5), r1=4+25−19=10r_1=\sqrt{4+25-19}=\sqrt{10}. Circle 2: x2+y2+2x+8y−23=0x^2+y^2+2x+8y-23=0, so g2=1,f2=4,c2=−23g_2=1,f_2=4,c_2=-23; centre C2=(−1,−4)C_2=(-1,-4), r2=1+16+23=40=210r_2=\sqrt{1+16+23}=\sqrt{40}=2\sqrt{10}. Distance C1C2=(2+1)2+(5+4)2=9+81=90=310=r1+r2=10+210C_1C_2=\sqrt{(2+1)^2+(5+4)^2}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}=r_1+r_2=\sqrt{10}+2\sqrt{10}. So the circles touch externally. …

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