If the tangents to the circle x^2 + y^2 = a^2 | vedclass

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(C) The equation of a tangent to the circle $x^2 + y^2 = a^2$ with slope $m$ is $y = mx \pm a\sqrt{1 + m^2}$.Let the slopes be $m_1 = 	an \alpha$ and

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$xy = 0$

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(C) The equation of a tangent to the circle $x^2 + y^2 = a^2$ with slope $m$ is $y = mx \pm a\sqrt{1 + m^2}$.Let the slopes be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$.The tangents are $y - m_1 x = \pm a\sqrt{1 + m_1^2}$ and $y - m_2 x = \pm a\sqrt{1 + m_2^2}$.Given $\cot \alpha + \cot \beta = 0$,which implies $\frac{1}{m_1} + \frac{1}{m_2} = 0$,so $m_1 + m_2 = 0$,or $m_2 = -m_1$.Let $m_1 = m$,then $m_2 = -m$.The tangents are $y - mx = \pm a\sqrt{1 + m^2}$ and $y + mx = \pm a\sqrt{1 + m^2}$.Multiplying these two equations: $(y - mx)(y + mx) = a^2(1 + m^2)$.$y^2 - m^2 x^2 = a^2 + a^2 m^2$.$y^2 - a^2 = m^2(x^2 + a^2)$.Since $m$ is a variable slope,for the intersection point $P(x, y)$ to exist for any such pair of tangents,the locus is found by eliminating $m$. However,given the condition $m_2 = -m_1$,the intersection point $(x, y)$ satisfies $x = 0$ or $y = 0$ depending on the configuration.Specifically,if $m_2 = -m_1$,the intersection point $P$ lies on the axes.Thus,the locus is $xy = 0$.

If the tangents to the circle $x^2 + y^2 = a^2$ with slopes $\alpha$ and $\beta$ intersect at point $P$,and $\cot \alpha + \cot \beta = 0$,then the locus of $P$ is:

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