The equation of the common tangent to the circles $x^2 + y^2 - 4x + 10y + 20 = 0$ and $x^2 + y^2 + 8x - 6y - 24 = 0$ is
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Whenever a question asks for a common tangent of two circles, start by checking $S_1 - S_2 = 0$. If that linear combination matches one of the options perfectly, it is almost certainly the correct tangent line at the contact interface!
Step 1: Read the two circles. $S_1:x^2+y^2-4x+10y+20=0$ and $S_2:x^2+y^2+8x-6y-24=0$. We find their centres and radii. Step 2: Centre and radius of $S_1$. Centre $C_1=(2,-5)$ and $r_1=\sqrt{2^2+(-5)^2-20}=\sqrt{4+25-20}=\sqrt9=3$. Step 3: Centre and radius of $S_2$. Centre $C_2=(-4,3)$ and $r_2=\sqrt{(-4)^2+3^2-(-24)}=\sqrt{16+9+24}=\sqrt{49}=7$. Step 4: Distance between centres. $C_1C_2=\sqrt{(2-(-4))^2+(-5-3)^2}=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=10$. Step 5: Test for external contact. Since $r_1+r_2=3+7=10=C_1C_2$, the circles touch externally, so the common tangent at the contact point is the radical axis $S_1-S_2=0$. Step 6: Subtract the equations. $S_1-S_2:(-4-8)x+(10+6)y+(20+24)=0$, i.e. $-12x+16y+44=0$. Dividing by $-4$ gives $3x-4y-11=0$, matching option (2). \[ \boxed{3x-4y-11=0} \]