Question

An equation of a circle touching the axes of coordinates and the line $x\cos \alpha + y\sin \alpha = 2$ is ${x^2} + {y^2} - 2gx + 2gy + {g^2} = 0$. Find the value of g.
A. $2{(\cos \alpha + \sin \alpha + 1)^{ - 1}}$
B. $2{(\cos \alpha - \sin \alpha + 1)^{ - 1}}$
C. $2{(\cos \alpha + \sin \alpha - 1)^{ - 1}}$
D. $-2{(\cos \alpha - \sin \alpha - 1)^{ - 1}}$

Answer 1

Hint: Write the center and the radius of the given circle. The line touches the coordinate axis and also the given line, hence apply that the perpendicular distance of the line from the center is the radius and calculate to obtain the value of g.

Formula Used:
The general equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + {r^2} = 0$ .
Here the center is $( - g, - f)$ and the radius is r.
The perpendicular distance of a line $ax + by + c = 0$ from a point $(p,q)$ is,
$d = \left| {\dfrac{{ap + bq + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$.

Complete step by step solution:
The given equation of the circle is ${x^2} + {y^2} - 2gx + 2gy + {g^2} = 0$.
Here the center is $(g, - g)$ and the radius is g.
The perpendicular distance of the line $x\cos \alpha + y\sin \alpha = 2$ from the center is the radius.
Therefore,
$\left| {\dfrac{{g\cos \alpha - g\sin \alpha - 2}}{{\sqrt {\cos {\alpha ^2} + {{( - \sin \alpha )}^2}} }}} \right| = g$
$g\cos \alpha - g\sin \alpha - 2 = \pm g$
$g(\cos \alpha - \sin \alpha \pm 1) = 2$
$g = \dfrac{2}{{\cos \alpha - \sin \alpha \pm 1}}$
That is,
$g = \dfrac{2}{{\cos \alpha - \sin \alpha + 1}},g = \dfrac{2}{{\cos \alpha - \sin \alpha - 1}}$.

Option ‘B’ is correct

Note: A circle touching the y−axis will have the absolute value of x−coordinate of the center equal to the length of the radius. To solve these types of questions we need to know some of the basic formulas to calculate the distance between two points