Question

The point on the curve $x^2+y^2-2x-3=0$ at which the tangent in parallel to x-axis is

• A. $(1,0),(-1,-4)$
• B. $(0,-1),(-2,3)$
• C. $(2,13),(-2,-3)$
• D. $(1,2),(1,-2)$

Answer 1

The correct option is D $(1,2),(1,-2)$

$(1,2)$ and $(1,-2)$ are two points where the tangent is parallel to the x-axis.

The equation of curve can be reduced to $(x-1{)}^2+y^2=4$

Therefore the center is $\left(1.0\right)$ on the x-axis and the radius is $2$ units

Now if we add and subtract $2$ to the $y$ co-ordinate of the center we get $(1,2)$ and $(1,-2)$ are two points where tangent is possible to x-axis.

$y=2$ and $y=-2$ are the two equations of the tangent to the circle and also parallel to axis and equation of the line$x=1$ represent diameter perpendicular to $x-axis$