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

Locus of point of intersection of tangents tothe circle x² + y2 = a w

venkat

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.

Last Activity: 7 Years ago

Namratha kris balram

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

Last Activity: 7 Years ago

venkat

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

and that is your answer.

Again sorry for my mistake.

Approved

Last Activity: 7 Years ago

Analytical Geometry> Locus of point of intersection of tangent...

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

Namratha kris balram , 7 Years ago

venkat

Namratha kris balram

venkat

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