Geo4.38

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Find the locus of point of intersection of two perpendicular tangents to the circle x^{2}+y^{2}=a^{2}\,

Given circle is

x^{2}+y^{2}=a^{2}\,

Let this equation be 1.

Let (x1,y1) be the point of intersection of two pependicular tangents.

Equation of a tangent to equation 1 is

y=mx+a{\sqrt  {(1+m^{2})}}\,

Let this equation be 2.

The equation 2 passes thro' (x1,y1), we have

y_{1}=mx_{1}+a{\sqrt  {(1+m^{2})}}\,

Simplifying

(y_{1}-mx_{1})^{2}=a^{2}(1+m^{2})\,

Rearranging

(x_{1}-a^{2})m^{2}-2x_{1}y_{1}m+(y_{1}^{{2}}-a^{2})=0\,

If m1,m2 are the two slopes of two tangents from (x1,y1) to 1,m1,m2 are the roots of the above quadratic equation,

Given two tangents are pependicular to each other, m1m2=-1

{\frac  {y_{1}^{{2}}-a^{2}}{x_{1}^{{2}}-a^{2}}}=-1\,

Hence

x_{1}^{{2}}+y_{1}^{{2}}=2a^{2}\,

Now the locus of the point (x1,y1) is

x^{2}+y^{2}=2a^{2}\,


Main Page:Geometry:Circles