# Locus of point of intersection of tangents tothe circle x² + y2 = a which makescomplementary angles with X-axis is1) x² y² = 0 2) x² + y² = 03) xy = 0 4) x² + y2 = 2a²

venkat
105 Points
4 years ago
If you understand the question clearly then you solve the above problem.
We know the equation of a tangent in terms of slope form
i.e.,$y=mx\pm r\sqrt{1+m^{2}}$
$y-mx=\pm r\sqrt{1+m^{2}}$
Squaring on both sides, we get
$(y-mx)^2=(\pm r\sqrt{1+m^{2}})^2$
$y^2+m^2x^2-2mxy=r^2(1+m^2)$
$y^2+m^2x^2-2mxy=r^2+r^2m^2$
$m^2(x^2-r^2)-2mxy+y^2-r^2=0$
For the lines to make complemetary angles with the x-axis the product of the roots of above equation is equal to -1
i.e.,,m1m2 = -1
$\Rightarrow m_1m_2=(y^2-r^2)/(x^2-r^2)=-1$
On simplifying you get the equation of a circle known as director circle
x2+y2=2a2 [since here r=a]
Approve my answer if it helped you.
Namratha kris balram
27 Points
4 years ago
M1 M2 = 1. Tan#*Cot #= 1 . So the tangent are not perpendicular to each other . The answer is x^2 - y ^2 =0 . But I don`t know how . Can you please explain the reason to me
venkat
105 Points
4 years ago
Oh i regret for my mistake. Please excuse me. I thought that the tangents were perpendicular.
But they are not perpendicular,but make complementary angles with the x-axis
i.e., if one of the tangents makes angle $\Theta$ with a slope say m1=tan$\Theta$
Then the other makes a complementary angle with axis (90-$\Theta$) with a slope tan(90-$\Theta$)=cot$\Theta$
Then clearly we know that tan$\Theta$.cot$\Theta$=1
i.e., m1m2=1
then, x2-r2=y2-r2