Question

Equation of the circle which touches the lines x = 0, y = 0 and 3x + 4y = 4 is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let centre of circle be (h, k). Since it touches both axes, therefore h = k = a
Hence equation can be $(x-a{)}^2+(y-a{)}^2=a^2$
But it also touches the line 3x + 4y = 4
Therefore, $\frac{3a+4a-4}{5}=|a|\Rightarrow a=2,1/3$
Hence the required equation of circle is
$x^2+y^2-4x-4y+4=0$