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Angle Bisector and It's Construction

A point P lie...

Question

A point P lies on the circle $x^{2} + y^{2} = 169$ . If $Q = (5, 12)$ and $R = (- 12, 5)$ , then the angle $∠ Q P R$ is

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is B $\frac{π}{4}$
$x^{2} + y^{2} = (13)^{2}$
radius $= 13$ center $(0, 0)$
Equation of $O Q = y = \frac{12}{5} x \Rightarrow 5 y - 12 x = 0$
Equation of $O R = y = \frac{- 5}{12} x \Rightarrow 12 y + 5 x = 0$
Angle between $O Q$ and $O R$
$tan θ = ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{12}{5} - \frac{5}{12}}{1 + (- 1)} ∣ ∣ ∣ ∣ ∣ ∣ = \infty θ = \frac{π}{2} ∠ Q P R = \frac{1}{2} ∠ Q O R = \frac{π}{4}$

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Similar questions

P

x^{2} + y^{2} = 169

Q = (5, 12)

R = (- 12, 5)

∠ Q P R

The $Δ P Q R$ is inscribed in the circle $x^{2} + y^{2} = 25$ .If Q and R have coordiantes (3,4) and (-4,3) respectively,then $∠ Q P R$ is equal to

x^{2} + y^{2} = 41

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

(\sqrt{3} sin θ, \sqrt{4} cos θ)

θ \in R

\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1

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x^{2} + y^{2} - 2 x - 2 y - 6 = 0

(\frac{π}{3}, π)

a

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