An equation of a circle touching the axes of coordinates and the line $x cos α + y sin α = 2$ can be

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

where

g = \frac{2}{(cos α + sin α + 1)}

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

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

where

g = \frac{2}{(cos α + sin α - 1)}

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

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

where

g = \frac{2}{(cos α - sin α + 1)}

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

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

where

g = \frac{2}{(cos α - sin α - 1)}

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

Open in App

Solution

The correct options are
A $x^{2} + y^{2} - 2 g x - 2 g y + g^{2} = 0$ where $g = \frac{2}{(cos α + sin α + 1)}$
B $x^{2} + y^{2} - 2 g x - 2 g y + g^{2} = 0$ where $g = \frac{2}{(cos α + sin α - 1)}$
C $x^{2} + y^{2} - 2 g x + 2 g y + g^{2} = 0$ where $g = \frac{2}{(cos α - sin α + 1)}$
D $x^{2} + y^{2} - 2 g x - 2 g y + g^{2} = 0$ where $g = \frac{2}{(cos α - sin α - 1)}$
Equation of a circle which touches the coordinate axes can be written as
$(x - g)^{2} + (y \pm g)^{2} = g^{2}, g$ being a variable.
If this circle touches the line $x cos α + y sin α = 2$
we get $g cos α \pm g sin α - 2 = \pm g \Rightarrow g = \frac{2}{cos α \pm sin α \pm 1}$
so the $(a), (b), (c)$ and $(d)$ are all correct.

Suggest Corrections

Similar questions

x cos α + y sin α = 2 i s x^{2} + y^{2} - 2 g x + 2 g y + g^{2} = 0

g =

Which of the following equations represents a circle with centre $(g, f)$ and radius $\sqrt{g^{2} + f^{2} + c}$ ?

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

If the circles $x^{2} + y^{2} = a^{2}$ and $x^{2} + y^{2} - 2 g x + g^{2} - b^{2} = 0$ touch each other externally, then

x^{2} + y^{2} + 2 g x + c = 0

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

g^{2} =

Join BYJU'S Learning Program