1
$\begingroup$

The angle between the pair of tangents drawn from a point P to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$ is $2\alpha$ . Then the equation of the locus of the point P is

A- $x^2+y^2+4x−6y+4=0$

B- $x^2+y^2+4x−6y-9=0$

C- $x^2+y^2+4x−6y-4=0$

D- $x^2+y^2+4x−6y+9=0$

My approach is as follow

${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0 \Rightarrow {\left( {x + 2} \right)^2} + {\left( {y - 3} \right)^2} = 13{\sin ^2}\alpha - 9{\sin ^2}\alpha = {\left( {2\left| {\sin \alpha } \right|} \right)^2} = {R^2}$

${X^2} + {Y^2} = {R^2}$

$Y = mX \pm \sqrt {{{\mathop{\rm R}\nolimits} ^2} + {R^2}{m^2}} $ is the general equation of the tangent

$\Rightarrow Y - mX = \pm R\sqrt {1 + {m^2}} \Rightarrow {Y^2} + {m^2}{X^2} - 2mYX = {R^2} + {R^2}{m^2}$

$ \Rightarrow {m^2}\left( {{X^2} - {R^2}} \right) - 2mYX + {Y^2} - {R^2} = 0$

$\tan 2\alpha = \frac{{{m_1} + {m_2}}}{{1 - {m_1}{m_2}}} = \frac{{\frac{{2YX}}{{\left( {{X^2} - {R^2}} \right)}}}}{{1 - \frac{{\left( {{Y^2} - {R^2}} \right)}}{{\left( {{X^2} - {R^2}} \right)}}}} = \frac{{2YX}}{{{X^2} - {Y^2}}}$

$2\sin \alpha = R \Rightarrow \sin \alpha = \frac{R}{2} \Rightarrow \tan \alpha = \frac{R}{{\sqrt {4 - {R^2}} }}$

$\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{{2YX}}{{{X^2} - {Y^2}}} = \frac{{2\left( {y - 3} \right)\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2} - {{\left( {y - 3} \right)}^2}}}$

not able to proceed from here

$\endgroup$

1 Answer 1

1
$\begingroup$

enter image description here

The choices should alert you that a simpler method is possible.

On drawing a diagram, we see that $P$ is at a constant distance from center $O=(-2,3)$. You can now finish.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.