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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. - Mathematics and Statistics

Sum

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.

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Solution

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

Alternate method:

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.

Concept: Tangent
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 6 Circle
Miscellaneous Exercise 6 | Q II. (25) | Page 138
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