Question

An equation of the tangent to the circle $x^2+y^2+4x-4y+4=0$ which makes equal intercepts on the coordinate axes, is given by

• A. $x+y=2\sqrt{2}$
• B. $x-y=2\sqrt{2}$
• C. $x+y+2\sqrt{2}=0$
• D. $x-y+2\sqrt{2}=0$

Answer 1

Answer: (A), (C)

The correct options are
A $x+y=2\sqrt{2}$
C $x+y+2\sqrt{2}=0$

Given equation of circle $x^2+y^2+4x-4y+4=0$ ....(1)
Center is at $(-2,2)$
Radius $r=\sqrt{4+4-4}=2$
Now, given tangent to eqn (1) makes equal intercepts on the coordinate axes,
So, let the equation of tangent be $x\pm y=a$
Now since, length of tangent from the center =Radius
$\Rightarrow \frac{|-2\pm 2-a|}{\sqrt{2}}=2$
Taking (+) sign , we get
$\Rightarrow a=\pm 2\sqrt{2}$
Hence equation of the required tangent is $x+y=\pm 2\sqrt{2}$