Question

Find the equations of direct common tangents for two circles

$x^2+y^2+6x-2y+1=0,$ $x^2+y^2-2x-6y+9=0$

• A. $35x^2+12xy-18x=0$
• B. $35y^2+12xy-18y=0$
• C. $3x^2-4xy+16y-12x=0$
• D. $3y^2-4xy+16x-12y=0$

Answer 1

The correct option is D

$3y^2-4xy+16x-12y=0$

Given circles,

$x^2+y^2+6x-2y+1=0$ - - - - - - (1)

$x^2+y^2-2x-6y+9=0$ - - - - - - (2)

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

$c_1(-3,1)$

$c_2(1,3)$

$r_1=\sqrt{g^2+f^2-c}=\sqrt{9+1-1}=3$

$r_2=\sqrt{g^2+f^2-c}=\sqrt{1+9-9}=1$

$c_1{c}_2=\sqrt{16+4}=\sqrt[2]{25}=4.47$

$c_1{c}_2>r_1+r_2$

Both the circle lies outside each other.

Let intersection point of direct common tangent be Q. Q divides $c_1\&c_2$

externally in the ratio of $r_1:r_2$ here, it divides in the ratio of 3:1. Co-ordinates of

Q(x,y)=$(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$=(\frac{3\times 1+1\times 1(-3)}{3-1},\frac{3\times 3-1\times \left(1\right)}{3-1})$

$=(\frac{6}{2},\frac{8}{2})=(3,4)$

Equation of direct common tangent or pair of tangent from the external point

$T^2=S{S}_1$

$T=x{x}_1+y{y}_1+3(x+x_1)-(y+y_1)+1$

$=3x+4y+3(x+3)-(y+4)+1$

$=6x+3y+6$

Substituting values of $T\&s_1$ in equation (3)

$(6x+3y+6{)}^2=(x^2+y^2+6x-2y+1\left)\right(3^2+4^2+18-8+1)$

$=(x^2+y^2+6x-2y+1)\left(36\right)$

$36x^2+9y^2+36+36xy+36y+72x=36x^2+36y^2+216x-72y+36$

$-27y^2+36xy-144x+108y=0$

$3y^2-4xy+16x-12y=0$

Equation of transverse common tangents

$3y^2+4xy+16x-12y=0$