Question

The equation of the circle which touches the axes of coordinates and the line \[\frac{x}{3} + \frac{y}{4} = 1\] and whose centres lie in the first quadrant is x² + y² − 2cx − 2cy + c² = 0, where c is equal to

विकल्प

• 4

• 2

• 3

• 6

Answer 1

6

The equation of the circle that touches the axes of coordinates is

\[x^2 + y^2 - 2cx - 2cy + c^2 = 0\]

Also, 

\[x^2 + y^2 - 2cx - 2cy + c^2 = 0\] touches the line 

\[\frac{x}{3} + \frac{y}{4} = 1\] or 4x + 3y \[-\] 12 = 0. 

Since the circle lies in the first quadrant, it centre is is \[\left( c, c \right)\].

 

From the figure, we have:

\[\left| \frac{4c + 3c - 12}{\sqrt{4^2 + 3^2}} \right| = c\]

\[ \Rightarrow \frac{7c - 12}{5} = c\]

\[ \Rightarrow c = 6\]