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Pair of Tangents from a Point

Let Q= a, b...

Question

Let $Q = (a, b)$ be a point. $P$ is a point on the circle centred at the origin and of radius $r$ . Let $α$ be the angle which the line joining $P$ to the centre makes with the positive x-axis. lf the line $P Q$ is a tangent to the circle, then acos $α +$ bsin $α =$

r

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r^{2}

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\frac{1}{r}

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\frac{1}{r^{2}}

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Solution

The correct option is A $r$
Equation of tangent to $x^{2} + y^{2} = r^{2}$ at P is
$x cos α + y sin α = r - - - - (1)$
So, $Q (a, b)$ is also passes through $x cos α + y sin α - r = 0$ , because PQ is tangent to circle.
So, $a cos α + b sin α - r = 0$
$\Rightarrow a cos α + b sin α = r$

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Let Q=(a,b) be a point. P is a point on the circle centred at the origin and of radius r. Let α be the angle which the line joining P to the centre makes with the positive x-axis. lf the line PQ is a tangent to the circle, then acos α+ bsin α=

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Let $Q = (a, b)$ be a point. $P$ is a point on the circle centred at the origin and of radius $r$ . Let $α$ be the angle which the line joining $P$ to the centre makes with the positive x-axis. lf the line $P Q$ is a tangent to the circle, then acos $α +$ bsin $α =$

The correct option is A $r$
Equation of tangent to $x^{2} + y^{2} = r^{2}$ at P is
$x cos α + y sin α = r - - - - (1)$
So, $Q (a, b)$ is also passes through $x cos α + y sin α - r = 0$ , because PQ is tangent to circle.
So, $a cos α + b sin α - r = 0$
$\Rightarrow a cos α + b sin α = r$