Question

$x^2+y^2+6x+8y=0$ and $x^2+y^2-4x-6y-12=0$ are the equation of the two circle Equation of one of their common tangent is

• A. $7x-5y-1-5\sqrt{74}=0$
• B. $7x-5y-1+5\sqrt{74}=0$
• C. $7x-5y+1-5\sqrt{74}=0$
• D. $5x-7y+1-5\sqrt{74}=0$

Answer 1

The correct option is C $7x-5y+1-5\sqrt{74}=0$

$C_1:x^2+y^2+6x+8y=0$

Radius $=\sqrt{3^2+4^2}=5$

$C_2:x^2+y^2-4x-6x-12=0$

Radius $=\sqrt{2^2+3^2+12}=5$

Distance between centre $=\sqrt{5^2+7^2}=\sqrt{74}$

$|r_1-r_2|<c_1{c}_2<r_1+r_2$

$2$ common tangent

Let tangent be $y=mx+c$

$\perp$ distance from $c_1$ to tangent $=5$

$\frac{|-4+3m-c|}{\sqrt{1+m^2}}$

$\perp$ distance from $c_2$ to tangent $=5$

$\frac{|3-2m-c|}{\sqrt{1+m^2}}$

$m=\frac{7}{5}$

$|-4+\frac{21}{5}-c|=5\times \frac{1}{5}\sqrt{74}$

$(-\frac{1}{5}-c)=\sqrt{74}$

$c=\frac{1}{5}-\sqrt{74},\sqrt{74}+\frac{1}{5}$

$mx+c-y=0\frac{7}{5}x+c-y=07x+5c-5y=0$

$7x-5y+1-5\sqrt{74}=0,7x-5y+1+5\sqrt{74}=0$

$C$ is correct.