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  • The locus of the feet of the perpendiculars drawn from the point (a, 0) on the tangents to the circle <img alt="\mathrm{x^2+y^2=a^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cma

The locus of the feet of the perpendiculars drawn from the point (a, 0) on the tangents to the circle \mathrm{x^2+y^2=a^2} is

Option: 1

\mathrm{\left(x^2-y^2-a x\right)^2=a^2\left[y^2+(x-a)^2\right]}


Option: 2

\mathrm{\left(x^2+y^2-a x\right)^2=a^2\left[(a-x)^2+y^2\right]}


Option: 3

\mathrm{(x-a)^2+y^2=a^2\left[x^2+y^2-a x\right]^2}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

Given circle is \mathrm{x^2+y^2=a^2} and equation of any tangent to it is

\mathrm{ y=m x+a \sqrt{1+m^2} }                                 ...(i)

Now equation of line through (a, 0) and perpendicular to (i) is given by

\mathrm{ y=\frac{a-x}{m} }                                                      ...(ii)
The locus of feet of \perp drawn from (a, 0) to the tangents of the circle will be same as the locus of point of intersection of (i) and (ii), so eliminating m from (i) and (ii), we have
\mathrm{ \begin{aligned} & y=\frac{(a-x) x}{y}+a \sqrt{1+\left(\frac{a-x}{y}\right)^2} \\ \Rightarrow & {\left[y^2-(a-x) x\right]^2=a^2\left[y^2+(a-x)^2\right] } \\ \Rightarrow & \left(x^2+y^2-a x\right)^2=a^2\left[y^2+(a-x)^2\right] \end{aligned} }

Posted by

HARSH KANKARIA

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