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
(a) 4
(b) 2
(c) 3
(d) 6

Answer 1

(d) 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:

$$\begin{gathered} \left|\frac{4c+3c-12}{\sqrt{4^2+3^2}}\right|=c \\ \Rightarrow \frac{7c-12}{5}=c \\ \Rightarrow c=6 \end{gathered}$$