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Rectangular Cartesian Coordinate System

Angle between...

Question

Angle between tangents drawn to $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ at the points where it is cut by the line $y = 2 x + c, i s \frac{π}{2}$ then

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Solution

The correct option is C
$x^{2} + y^{2} - 20 - 4 y + 1 = 0 y = 20 + c$
Radius of the circle = 2
perpendicular distance of line y = 20 + c from the centre of the circle, $d = \frac{c}{\sqrt{5}}$
$c o s θ = \frac{d}{r}, (θ = \frac{π}{4}) \frac{c}{2 \sqrt{5}} = \frac{1}{\sqrt{2}} c = \sqrt{10}$

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Similar questions

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

y = 2 x + c, i s \frac{π}{2}

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

y = 2 x + c

\frac{π}{2}

x^{2} + y^{2} + 4 x + 8 y + c = 0

(0, 0)

\frac{π}{2}

^{'} c^{'}

c

y = c

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

(1, 1)

C

\sqrt{10}

y = 2 x + \frac{5}{\sqrt{2}}

45^{\circ}

C

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