Question

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

• A. $x+y=2$
• B. $x+y=2\sqrt{2}$
• C. $x+y=4$
• D. $x+y=8$

Answer 1

Answer: (B)

$x+y=2\sqrt{2}$

Equation of a line which makes equal intercepts on the positive coordinate axes is given by,
$x+y=a$, $a>0$
So, $x+y=a$ is a tangent to $x^2+y^2+4x-4y+4=0$

$\therefore$ Perpendicular distance of centre $(-2,2)$ from line $x+y=a$ is equal to radius of the circle.

Perpendicular distance of centre $(x_1,y_1)$ from line $ax+by+c=0$ is

$\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

Here, Radius $=2=\left|\frac{-2+2-a}{\sqrt{2}}\right|$
$\Rightarrow a=2\sqrt{2}$

Therefore equation of tangent is
$\Rightarrow x+y=2\sqrt{2}$

Hence, option B.