The equation of the circle whose centre is 3-4 and radius is 5 is

The correct option is A r^2 6rcos( θ π4)=16 Centre is #160; (3, #160;π/4) General equation of circle in polar form is r ^ 2 2r r _ 0 co ...

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Standard XII

Physics

Introduction

The equation ...

Question

The equation of the circle whose centre is $(3, \frac{π}{4})$ and radius is $5$ is

r^{2} - 6 r cos (θ - \frac{π}{4}) = 16

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r^{2} + 6 r cos (θ - \frac{π}{4}) = 16

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r^{2} - 6 r cos (θ - \frac{π}{4}) + 16 = 0

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r^{2} - 8 r cos (θ - \frac{π}{6}) = 9

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Solution

The correct option is A $r^{2} - 6 r cos (θ - \frac{π}{4}) = 16$
Centre is $(3, \frac{π}{4})$
General equation of circle in polar form is $r^{2} - 2 r r_{0} cos (θ - γ) + {r_{0}}^{2} = a^{2}$
where, $(r_{0}, γ)$ is center and $a$ is radius
$\Rightarrow r^{2} - 2 r (3) cos (θ - \frac{π}{4}) + 3^{2} = 5^{2}$
$\Rightarrow r^{2} - 6 r cos (θ - \frac{π}{4}) = 16$
Ans: A

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Similar questions

Q. The centre of the circle

r^{2} - 8 r cos (θ - \frac{π}{6}) = 9

Q. The radius of the circle

r^{2} - 8 r cos (θ - \frac{π}{3}) = 9

Solve the equation: $\sqrt{(\frac{1}{16} + c o s^{4} x - \frac{1}{2} c o s^{2} x)} + \sqrt{(\frac{9}{16} + c o s^{4} x - \frac{3}{2} c o s^{2} x)} = \frac{1}{2}$

Q. A =

π r^{2}

, rewrite the formula '

π

' as a subject.

Let $z$ be a complex number such that $∣ ∣ ∣ z + \frac{1}{z} ∣ ∣ ∣ = 2$ .

If $| z | = r_{1}$ and $r_{2}$ for $arg z = \frac{π}{4}$ then

$| r_{1} - r_{2} | =$

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The equation of the circle whose centre is (3,π4) and radius is 5 is

The correct option is A r2−6rcos(θ−π4)=16Centre is (3,π4)General equation of circle in polar form is r2−2rr0cos(θ−γ)+r02=a2where, (r0,γ) is center and a is radius⇒r2−2r(3)cos(θ−π4)+32=52⇒r2−6rcos(θ−π4)=16Ans: A

The equation of the circle whose centre is $(3, \frac{π}{4})$ and radius is $5$ is