Question

Locus of points from which perpendicular tangents can be drawn to the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] is
(a) A circle passing through origin
(b) A circle of radius \[2a\]
(c) Concentric circle of radius \[a\sqrt{2}\]
(d) None of these.

Answer 1

Hint: For solving this problem we assume that the point from which the tangents are drawn as \[\left( h,k \right)\] and we take the equation of tangent as general form of equation that is \[y=mx+c\] where \['m'\] is a slope of tangent and \['c'\] is some constant. By substituting the line equation we get a value of \['c'\]. By substituting the point in line equation we get a value of \['c'\]. By equating both the values of \['c'\] we get the quadratic equation in \['m'\] since the tangents are perpendicular by taking the product of slopes as ‘-1’ we get the required locus.

Complete step-by-step solution

Let us assume that the point from where the tangents are drawn as \[P\left( h,k \right)\]
Let us assume that the equation of tangent as \[y=mx+c\] where \['m'\] is the slope of the tangent and \['c'\] is some constant.
Here, we know that the tangent is passing through \[P\left( h,k \right)\].
So, \[P\left( h,k \right)\] satisfies the tangent equation. By substituting \[P\left( h,k \right)\] in tangent equation we get
\[\begin{align}
  & \Rightarrow y=mx+c \\
 & \Rightarrow k=mh+c \\
 & \Rightarrow c=mh-k......equation(i) \\
\end{align}\]
We know that the tangent and the circle will intersect. So, let us substitute the value of \[y=mx+c\] in circle equation given as \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{\left( mx+c \right)}^{2}}={{a}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{m}^{2}}{{x}^{2}}+2mxc+{{c}^{2}}={{a}^{2}} \\
 & \Rightarrow {{x}^{2}}\left( 1+{{m}^{2}} \right)+x\left( 2mc \right)+\left( {{c}^{2}}-{{a}^{2}} \right)=0 \\
\end{align}\]
Since, we know that the tangent touches the circle, the discriminant of the above equation should be zero.
The discriminant of quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as \[{{b}^{2}}-4ac\].
By comparing the above equation with general equation of quadratic equation we get
\[\begin{align}
  & \Rightarrow {{\left( 2mc \right)}^{2}}-4\left( {{c}^{2}}-{{a}^{2}} \right)\left( 1+{{m}^{2}} \right)=0 \\
 & \Rightarrow 4{{m}^{2}}{{c}^{2}}-4{{c}^{2}}\left( 1+{{m}^{2}} \right)+4{{a}^{2}}\left( 1+{{m}^{2}} \right)=0 \\
 & \Rightarrow {{c}^{2}}\left( -1 \right)+{{a}^{2}}\left( 1+{{m}^{2}} \right)=0 \\
 & \Rightarrow {{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right)....equation(ii) \\
\end{align}\]
By substituting the value of \['c'\] we got from equation (i) in equation (ii) we get
\[\begin{align}
  & \Rightarrow {{\left( mh-k \right)}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right) \\
 & \Rightarrow {{m}^{2}}{{h}^{2}}-2mhk+{{k}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}} \\
 & \Rightarrow {{m}^{2}}\left( {{h}^{2}}-{{a}^{2}} \right)-2mhk+\left( {{k}^{2}}-{{a}^{2}} \right)=0....equation(iii) \\
\end{align}\]
We know that from the above equation we get two values of \['m'\] which gives two equations of tangents.
We are given that the tangents are perpendicular so the product of two slopes is ‘-1’.
For the general quadratic equation \[a{{x}^{2}}+bx+c=0\] the product of roots is given as\[\dfrac{c}{a}\].
By applying the product of roots to equation (iii) we get
\[\begin{align}
  & \Rightarrow \dfrac{{{k}^{2}}-{{a}^{2}}}{{{h}^{2}}-{{a}^{2}}}=-1 \\
 & \Rightarrow {{k}^{2}}-{{a}^{2}}=-{{h}^{2}}+{{a}^{2}} \\
 & \Rightarrow {{h}^{2}}+{{k}^{2}}={{\left( a\sqrt{2} \right)}^{2}} \\
\end{align}\]
By replacing \[\left( h,k \right)\] by \[\left( x,y \right)\] we get locus of point \[P\left( h,k \right)\] as
\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{\left( a\sqrt{2} \right)}^{2}}\]
Therefore, the required locus is a circle with radius \[a\sqrt{2}\].
So, option (c) is the correct answer.

Note: This problem can be solved in other methods also.
We know that the equation of tangent to general equation of circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] is given as
\[y=mx\pm a\sqrt{1+{{m}^{2}}}\]. We also know that point \[P\left( h,k \right)\] lies on the tangent we can write
\[\begin{align}
  & \Rightarrow k=mh\pm a\sqrt{1+{{m}^{2}}} \\
 & \Rightarrow k-mh=\pm a\sqrt{1+{{m}^{2}}} \\
\end{align}\]
By squaring on both sides we get
\[\begin{align}
  & \Rightarrow {{\left( mh-k \right)}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right) \\
 & \Rightarrow {{m}^{2}}{{h}^{2}}-2mhk+{{k}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}} \\
 & \Rightarrow {{m}^{2}}\left( {{h}^{2}}-{{a}^{2}} \right)-2mhk+\left( {{k}^{2}}-{{a}^{2}} \right)=0....equation(iii) \\
\end{align}\]
We know that from the above equation we get two values of \['m'\] which gives two equations of tangents.
We are given that the tangents are perpendicular so the product of two slopes is ‘-1’.
For the general quadratic equation \[a{{x}^{2}}+bx+c=0\] the product of roots is given as\[\dfrac{c}{a}\].
By applying the product of roots to equation (iii) we get
\[\begin{align}
  & \Rightarrow \dfrac{{{k}^{2}}-{{a}^{2}}}{{{h}^{2}}-{{a}^{2}}}=-1 \\
 & \Rightarrow {{k}^{2}}-{{a}^{2}}=-{{h}^{2}}+{{a}^{2}} \\
 & \Rightarrow {{h}^{2}}+{{k}^{2}}={{\left( a\sqrt{2} \right)}^{2}} \\
\end{align}\]
By replacing \[\left( h,k \right)\] by \[\left( x,y \right)\] we get locus of point \[P\left( h,k \right)\] as
\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{\left( a\sqrt{2} \right)}^{2}}\]
Therefore, the required locus is a circle with radius \[a\sqrt{2}\].
So, option (c) is the correct answer.