Q45 / 90 · JEE Main 2012 (26 May Online) · Mathematics

The number of common tangents of the circles given by $x^2+y^2-8 x-2 y+1=0$ and $x^2+y^2+6 x+8 y=0$ is

A
one
B
four
Correct answer
C
two
D
three

Solution

Given circles are $x^2+y^2-8 x-2 y+1=0$ and $x^2+y^2+6 x+8 y=0$ Their centres and radius are $$ C_1(4,1), r_1=\sqrt{16}=4 $$ $$ C_2(-3,-4), r_2=\sqrt{25}=5 $$ Now, $C_1 C_2=\sqrt{49+25}=\sqrt{74}$ $$ r_1-r_2=-1, r_1+r_2=9 $$ Since, $r_1-r_2 < C_1 C_2 < r_1+r_2$ $\therefore$ Number of common tangents $=2$

Hence, the correct answer is option: 'C'.

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