Question

If $y=c$ is a tangent to the circle $x^2+y^2-2x+2y-2=0$, then the value of c can be

• A. $1$
• B. $3$
• C. $-1$
• D. $-3$

Answer 1

Answer: (A), (D)

The correct options are
A $1$
D $-3$

Given, the equation of circle is:

$x^2+y^2-2x+2y-2=0$

$\Rightarrow x^2-2x+1+y^2+2y+1=4$

$\Rightarrow {(x-1)}^2+{(y+1)}^2=2^2$

Therefore, the circle has centre at $(1,-1)$ and radius of $2$ units.

Now, given that $y=c$ is a tangent to the circle.

Therefore, the distance between centre and tangent to circle must be equal to its radius.

Distance between $y=c$ and $(1,-1)$ must be equal to 2.

$\Rightarrow |c-(-1\left)\right|=2\Rightarrow |c+1|=2\Rightarrow c+1=2,-2\Rightarrow c=1,-3$

Hence, Options $\left(A\right)$ and $\left(D\right)$ are correct.