If {theta _1} and {theta _2} are the inclinat | vedclass

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(C) The equation of any tangent to the circle ${x^2} + {y^2} = {a^2}$ with slope $m$ is $y = mx \pm a\sqrt{1 + {m^2}}$.If this tangent passes through

Vedclass · Accepted Answer

$c({y^2} - {a^2}) = 2xy$

Vedclass · Answer

(C) The equation of any tangent to the circle ${x^2} + {y^2} = {a^2}$ with slope $m$ is $y = mx \pm a\sqrt{1 + {m^2}}$.If this tangent passes through the point $P(h, k)$,then $(k - mh)^2 = {a^2}(1 + {m^2})$.Expanding this,we get ${k^2} - 2mhk + {m^2}{h^2} = {a^2} + {a^2}{m^2}$,which rearranges to ${m^2}({h^2} - {a^2}) - 2mhk + ({k^2} - {a^2}) = 0$.Since $m = 	an 	heta$,the roots of this quadratic equation in $m$ are ${m_1} = 	an {	heta _1}$ and ${m_2} = 	an {	heta _2}$.Given $\cot {	heta _1} + \cot {	heta _2} = c$,we have $\frac{1}{{{m_1}}} + \frac{1}{{{m_2}}} = c$,which implies $\frac{{{m_1} + {m_2}}}{{{m_1}{m_2}}} = c$.From the quadratic equation,the sum of roots ${m_1} + {m_2} = \frac{2hk}{{{h^2} - {a^2}}}$ and the product of roots ${m_1}{m_2} = \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}}$.Substituting these into the expression,we get $\frac{2hk}{{{k^2} - {a^2}}} = c$,which simplifies to $c({k^2} - {a^2}) = 2hk$.Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $c({y^2} - {a^2}) = 2xy$.

If ${\theta _1}$ and ${\theta _2}$ are the inclinations of the tangents drawn from a point $P(h, k)$ to the circle ${x^2} + {y^2} = {a^2}$ with the $x$-axis,then the locus of $P$,given that $\cot {\theta _1} + \cot {\theta _2} = c$,is:

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