Question

Find the locus of the point of intersection of perpendicular tangents to the circle \[S \equiv {x^2} + {y^2} - 2x + 2y - 2 = 0\], that is, find the director circle of \[S = 0\].

Answer 1

Hint: First of all, find the centre of the circle and radius of the given circle. Then find the equation of the director circle as the director circle is a concentric circle whose radius is \[\sqrt 2 \] times the radius of the given circle.
Complete step-by-step answer:
The equation of the circle is \[S \equiv {x^2} + {y^2} - 2x + 2y - 2 = 0\].
We know that for the circle equation \[S \equiv a{x^2} + b{y^2} + 2gx + 2fy + c = 0\], centre of the circle is \[\left( { - g, - f} \right)\] and radius is \[\sqrt {{g^2} + {f^2} - c} \].
If we compare the given circle equation i.e., \[S \equiv {x^2} + {y^2} - 2x + 2y - 2 = 0\] with the standard circle equation \[S \equiv a{x^2} + b{y^2} + 2gx + 2fy + c = 0\], we have \[g = - 1\] and \[f = 1\].
So, the centre of the given circle is \[\left( {1, - 1} \right)\] and radius is\[\sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} - \left( { - 2} \right)} = \sqrt 4 = 2\].

We know that the director circle is a concentric circle whose radius is \[\sqrt 2 \] times the radius of the given circle. And the equation of the circle with centre \[\left( {h,k} \right)\] and radius \[r\] is given by \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\].
So, the equation of the director circle with centre \[\left( {1, - 1} \right)\] and radius \[2\sqrt 2 \] is given by
\[
   \Rightarrow {\left( {x - 1} \right)^2} + {\left( {y - \left( { - 1} \right)} \right)^2} = {\left( {2\sqrt 2 } \right)^2} \\
   \Rightarrow {x^2} - 2x + 1 + {\left( {y + 1} \right)^2} = {\left( 2 \right)^2}{\left( {\sqrt 2 } \right)^2} \\
   \Rightarrow {x^2} - 2x + 1 + {y^2} + 2y + 1 = 8 \\
   \Rightarrow {x^2} + {y^2} - 2x + 2y + 1 + 1 - 8 = 0 \\
  \therefore {x^2} + {y^2} - 2x + 2y - 6 = 0 \\
\]
Hence, the equation of the director circle is \[S = {x^2} + {y^2} - 2x + 2y - 6 = 0\].
Thus, the locus of the point of intersection of perpendicular tangents to the circle \[S \equiv {x^2} + {y^2} - 2x + 2y - 2 = 0\], that is, the director circle is \[S = {x^2} + {y^2} - 2x + 2y - 6 = 0\].
Note: The locus of points of intersection of two perpendicular tangents to a circle is called the director circle. For the circle equation \[S \equiv a{x^2} + b{y^2} + 2gx + 2fy + c = 0\], centre of the circle is \[\left( { - g, - f} \right)\] and radius is \[\sqrt {{g^2} + {f^2} - c} \].