Question

The equation of the circle which touches x the axes of coordinates and the line $\frac{x}{3}+\frac{y}{4}=1$ and whose centres lie in the first quadrant is $x^2+y^2-2cx-2cy+c^2=0$ where c is equal to

• A. 4
• B. 2
• C. 3
• D. 6

Answer 1

The correct option is 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 x line
$\frac{x}{3}+\frac{y}{4}=1or4x+3y-12=0$.

Since the circle lies in the first quadrant. It centre is (c, c)

From the figure, we have:

$\left|\begin{array}{c}\frac{4c+3c-12}{\sqrt{4^2+3^2}}=c\end{array}\right|$

$\Rightarrow \frac{7c-12}{5}=c$

$\Rightarrow c=6$