Question

The equation of common tangent to the circles $x^2+y^2=4$ and $x^2+y^2-6x-8y-24=0$ is

• A. $3x-4y+10=0$
• B. $3x-4y-10=0$
• C. $3x+4y+10=0$
• D. $3x+4y-10=0$

Answer 1

The correct option is C $3x+4y+10=0$

Given circles are
$x^2+y^2=4\cdots \left(1\right)$
$C_1=(0,0),r_1=2$
and
$x^2+y^2-6x-8y-24=0\cdots \left(2\right)$
$C_2=(3,4),r_2=7$
Now,
$C_1{C}_2=5=r_2-r_1$
So, two circles touch each other internally.
Let the equation of the tangent be $y=mx+c$
Common tangent is perpendicular to line joining centres, so
$m=-\frac{3-0}{4-0}=-\frac{3}{4}$
Tangent is $3x+4y-c=0$
Now, distance from centre to tangent is equal to radius,
$\frac{|-c|}{5}=2\Rightarrow c=\pm 10\frac{|25-c|}{5}=7\Rightarrow 25-c=\pm 35\Rightarrow c=-10,60\therefore c=-10$

Hence, the equation of tangent is
$3x+4y+10=0$