The locus of the point of intersection of the | vedclass

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(A) Let the slopes of the two tangents be $m_1$ and $m_2$. Since they make complementary angles with the $X$-axis,we have $m_1 = 	an(	heta)$ and $m_

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$x^2-y^2=0$

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(A) Let the slopes of the two tangents be $m_1$ and $m_2$. Since they make complementary angles with the $X$-axis,we have $m_1 = 	an(	heta)$ and $m_2 = 	an(90^\circ - 	heta) = \cot(	heta)$.Thus,$m_1 m_2 = 	an(	heta) \cdot \cot(	heta) = 1$.The equation of a tangent to the circle $x^2+y^2=a^2$ with slope $m$ is $y = mx \pm a\sqrt{1+m^2}$.Rearranging gives $y - mx = \pm a\sqrt{1+m^2}$.Squaring both sides,$(y-mx)^2 = a^2(1+m^2)$,which simplifies to $m^2(x^2-a^2) - 2mxy + (y^2-a^2) = 0$.This is a quadratic in $m$,where $m_1$ and $m_2$ are the roots.For the product of roots $m_1 m_2 = 1$,we use the property of quadratic equations $Ax^2+Bx+C=0$ where the product of roots is $C/A$.Here,$C = y^2-a^2$ and $A = x^2-a^2$.So,$\frac{y^2-a^2}{x^2-a^2} = 1$.$y^2-a^2 = x^2-a^2$,which implies $x^2-y^2=0$.

The locus of the point of intersection of the tangents to the circle $x^2+y^2=a^2$ which make complementary angles with the $X$-axis is

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