Question

If the equation of the circle lying in the first quadrant, touching both the coordinate axes and the line \( \frac{x}{3} + \frac{y}{4} = 1 \) is \( (x-c)^2+(y-c)^2=c^2 \), then c =

• A. 1 or 4
• B. 2 or 3
• C. 1 or 6
• D. 2 or 5

Hint:

A circle touching both axes in the 1st quadrant has centre \((c,c)\) and radius \(c\), so equation is \((x-c)^2+(y-c)^2=c^2\). The condition for a line to touch a circle is that the perpendicular distance from the centre to the line equals the radius. Distance from \( (x_0,y_0) \) to \( Ax+By+C=0 \) is \( \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} \).

Answer 1

The Correct Option is C

A circle in the first quadrant touching both axes has centre \( (c,c) \) and radius \(c\) (for \(c>0\)).
The equation is \( (x-c)^2 + (y-c)^2 = c^2 \).
The line is \( \frac{x}{3} + \frac{y}{4} = 1 \), which is \( 4x+3y=12 \), or \( 4x+3y-12=0 \).
Since the circle touches this line, the perpendicular distance from centre \( (c,c) \) to the line \( 4x+3y-12=0 \) equals the radius \(c\).
\[ \text{Distance} = \frac{|4(c)+3(c)-12|}{\sqrt{4^2+3^2}} = c \] \[ \frac{|7c-12|}{\sqrt{16+9}} = c \implies \frac{|7c-12|}{5} = c \] \[ |7c-12| = 5c \] This yields two cases: Case 1: \( 7c-12 = 5c \) \[ 2c = 12 \implies c=6 \] Case 2: \( 7c-12 = -5c \) \[ 12c = 12 \implies c=1 \] Both \(c=1\) and \(c=6\) are positive.
The possible values for c are 1 or 6.
This matches option (3).