Question

Find the length of direct common tangent for circles $x^2+y^2-20x+64=0$and $x^2+y^2+30x+144=0$

• A. $2\sqrt{150}$
• B. 20
• C. $2\sqrt{154}$
• D. 9

Answer 1

The correct option is C

$2\sqrt{154}$

Given circles

$x^2+y^2-20x+64=0$ - - - - - - (1)

$x^2+y^2+30x+144=0$ - - - - - - (2)

Let the circle be $c_1\&c_2$ and radii $r_1\&r_2$ of circle (1) and (2) respectively.

$c_1(10,0)$ $c_2(-15,0)$

$c_1{c}_2=\sqrt{(25{)}^2+0}=25$

$r_1=\sqrt{g^2+f^2-c}=\sqrt{100+0+64}=6$

$r_2=\sqrt{g^2+f^2-c}=\sqrt{225+0-144}=9$

$c_1{c}_2>r_1+r_2$

Both the circle neither touch nor intersect to each other.

Length of direct common tangent

$=\sqrt{d^2-(r_1-r_2{)}^2}$

$=\sqrt{(25{)}^2-(6-9{)}^2}$

$=\sqrt{625-9}=\sqrt{616}=2\sqrt{154}$

Length of direct common tangent =$2\sqrt{154}$units