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Question
Answer the following :
Find the equation of the locus of a point, the tangents from which to the circle x2 + y2 = 9 are at right angles.
Solution 1
Given the equation of the circle is
x2 + y2 = 9
Comparing this equation with x2 + y2 = a2, we get
a2 = 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 x2 + y2 = a2 is x2 + y2 = 2a2.
∴ the required equation is
x2 + y2 = 2(9)
∴ x2 + y2 = 18
Solution 2
Given equation of the circle is
x2 + y2 = 9
Comparing this equation with x2 + y2 = a2, we get
a2 = 9
Let P(x1, y1) be a point on the required locus. Equations of the tangents to the circle x2 + y2 = a2 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 (x1, y1).
∴ y1 = `"m"x_1 ± 3sqrt(1 + "m"^2)`
∴ y1 – mx1 = `± 3sqrt(1 + "m"^2)`
∴ (y1 – mx1)2 = 9(1 + m2) …[Squaring both the sides]
∴ `"y"_1^2` – 2mx1y1 + `"m"^2"x"_1^2` = 9 + 9m2
∴ (`"x"_1^2` – 9)m2 – 2mx1y1 + (`"y"_1^2` – 9) = 0
This is a quadratic equation which has two roots m1 and m2.
∴ m1m2 = `(y_1^2 - 9)/(x_1^2 - 9)`
Since, the tangents are at right angles.
∴ m1m2 = – 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 x2 + y2 = 18.
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