Answer the following :

Find the equation of the locus of a point, the tangents from which to the circle x^{2} + y^{2} = 9 are at right angles.

#### Solution

Given the equation of the circle is

x^{2} + y^{2} = 9

Comparing this equation with x^{2} + y^{2} = a^{2}, we get

a^{2} = 9

The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.

The equation of the director circle of the circle x^{2} + y^{2} = a^{2} is x^{2} + y^{2} = 2a^{2}.

∴ the required equation is

x^{2} + y^{2} = 2(9)

∴ x^{2} + y^{2} = 18

**Alternate method:**

Given equation of the circle is

x^{2} + y^{2} = 9

Comparing this equation with x^{2} + y^{2} = a^{2}, we get

a^{2} = 9

Let P(x_{1}, y_{1}) be a point on the required locus. Equations of the tangents to the circle x^{2} + y^{2} = a^{2} with slope m are

y = `"m"x ± sqrt("a"^2 (1 + "m"^2)`

∴ Equations of the tangents are

y = `"m"x ± sqrt(9("m"^2 + 1)`

∴ y = `"m"x ± 3sqrt(1 + "m"^2)`

Since, these tangents pass through (x_{1}, y_{1}).

∴ y_{1} = `"m"x_1 ± 3sqrt(1 + "m"^2)`

∴ y_{1} – mx_{1} = `± 3sqrt(1 + "m"^2)`

∴ (y_{1} – mx_{1})^{2} = 9(1 + m^{2}) …[Squaring both the sides]

∴ `"y"_1^2`^{ } – 2mx_{1}y_{1} + `"m"^2"x"_1^2` = 9 + 9m^{2}

∴ (`"x"_1^2` – 9)m^{2} – 2mx_{1}y_{1} + (`"y"_1^2` – 9) = 0

This is a quadratic equation which has two roots m_{1} and m_{2}.

∴ m_{1}m_{2} = `(y_1^2 - 9)/(x_1^2 - 9)`

Since, the tangents are at right angles.

∴ m_{1}m_{2} = – 1

∴ `(y_1^2 - 9)/(x_1^2 - 9)` = – 1

∴ `"y"_1^2` – 9 = 9 – `"x"_1^2`

∴`"x"_1^2+"y"_1^2` = 18

∴ Equation of the locus of point P is x^{2} + y^{2} = 18.