Question

A circle having its centre at (2, 3) is cut orthogonally by the parabola $y^2=4x$. The possible intersection point(s) of these curves, can be

• A. $(3,2\sqrt{3})$
• B. $(2,2\sqrt{3})$
• C. (1,2)
  
• D. (4,3)

Answer 1

The correct option is C

(1,2)

Any tangent to the parabola $y^2=4xat(t^2,2t)$ is $yt=x+t^2$ If it passes through the centre (2, 3) of the circle, then
$t^2-3t+2=0\Rightarrow t=1,2$
$\therefore$ The point can be (1, 2) or (4, 4)