Question

The equation of the pair of tangent drawn to the circle $x^2+y^2-2x+4y+3=0$ from $(6,-5)$ is

• A. $7x^2+23y^2+30xy+66x+50y-73=0$
• B. $7x^2+23y^2-30xy-66x-50y+73=0$
• C. $7x^2+23y^2+30xy-66x-50y-73=0$
• D. $Noneoftheabove$

Answer 1

The correct option is A $7x^2+23y^2+30xy+66x+50y-73=0$

given circle equation is $x^2+y^2-2x+4y+3=0$

This can be written as $(x-1{)}^2+(y+2{)}^2=2$

Let the equation of the line be $y=mx+c$

It passes through $(6,-5)$ so, $c=-(5+6m)$

equation of the line is $y=mx-(5+6m)$

substituting line equation in the circle equation, $(x-1{)}^2+(mx+2-(5+6m){)}^2=0$

expanding and rearranging we get,

$(1+m^2)x^2-x(12m^2+6m+2)+(36m^2+36m+8)=0$

This equation a unique solution, because the line is a tanget to the given circle.

$b^2=4ac$

$(12m^2+6m+2{)}^2=4(1+m^2\left)\right(36m^2+36m+8)$

rearranging the terms we get, $23m^2+30m+7=\mathrm{0....}\left(1\right)$

$m_1+m_2=\frac{-30}{23},m_1{m}_2=\frac{7}{23}$

let the line equations be $y=m_{1}x+c_1,y=m_{2}x+c_2$

$c_1=-(5+6m_1),c_2=-(5+6m_2)$

equation of pair of lines is $(m_{1}x-y+c_1)(m_{2}x-y+c_2)=0$

$=m_1{m}_2{x}^2+y^2+(m_1+m_2)xy+(m_1{c}_2+c_1{m}_2)x-y(c_1+c_2)+c_1{c}_2=0$

From equation (1), $m_1+m_2=\frac{-30}{23},m_1{m}_2=\frac{7}{23},c_1+c_2=-(10+12(m_1+m_2\left)\right)=\frac{-50}{23}$

$c_1{m}_2+m_1{c}_2=-\left(5\right(m_1+m_2)+12m_1{m}_2)=\frac{66}{23}$

$c_1{c}_2=(5+6m_1)(5+6m_2)=25+30(m_1+m_2)+36m_1{m}_2=\frac{-73}{23}$

Substituting corresponding values we get,

$7x^2+23y^2+30xy+66x+50y-73=0$