Question

Find the equations of transverse 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 A

$35x^2+12xy-18x=0$

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

$c_1(-3,1)$

$c_2(1,3)$

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

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

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

$c_1{c}_2>r_1+r_2$

Both the circle lies outside each other.

Let the intersection point of tarnsverse common tangent be P. P divides $c_1\&c_2$ internelly in the ratioo of

$r_1:r_2$ here ,in the ratio of 3:1.Co-ordinate of P(x,y)=$(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$=(\frac{3\times 1+1\times 1-3}{3+4},\frac{3\times 3+1\times 1}{3+1})$

$=(0,\frac{5}{2})$

Equation of transvers common tangent or pair of tangent from the point P is

$T^2=S{S}_1$ - - - - -(3)

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

$=x\times 0_y\frac{5}{2}+3(x+0)-(y+\frac{5}{2})+1$

$=\frac{5}{2}y+3x-y-\frac{5}{2}+1=\frac{3}{2}y+3x-\frac{3}{2}$

$\frac{3}{2}(y+6x-1)$

$s_1=(0{)}^2+{\left(\frac{5}{2}\right)}^2+6(0)-2\left(\frac{5}{2}\right)+1$

$=\frac{25}{4}-\frac{10}{2}+1=\frac{25-20+4}{4}=\frac{9}{4}$

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

$\frac{9}{4}(y+6x-1)2=(x^2+y^2+6x-2y+1)\times \frac{9}{4}$

$35x^2+12xy-18x=0$

Equation of transverse common tangents

$35x^2+12xy-18x=0$