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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $d = \frac{c}{\sqrt{5}}$
$c o s θ = \frac{d}{r}, (θ = \frac{π}{4}) \frac{c}{2 \sqrt{5}} = \frac{1}{\sqrt{2}} c = \sqrt{10}$

Suggest Corrections

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} - 2 x - 4 y + 1 = 0

y = 2 x + c, i s \frac{π}{2}

x^{2} + y^{2} + 2 x + 4 y - 31 = 0

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

(4, 3)

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

Join BYJU'S Learning Program