Question

The locus of the center of a circle which cuts orthogonally the parabola $y^2=4x$ at $(1,2)$ will pass through.

• A. $(3,4)$
• B. $(4,3)$
• C. $(5,3)$
• D. $(2,4)$

Answer 1

The correct option is A $(3,4)$

let the center of circle be $(h,k)$ and radius be $r$, equation of circle be, $=>(x-h{)}^2+{(y-k)}^2=r^2......(1)$

As it cuts parabola at $(1,2)$ it will pass through it,

$=>(1-h{)}^2+{(2-k)}^2=r^2.............(2)$

Given Parabola: { y }^{ 2 }=4x\quad (a=1)\quad ...........(3)$

Tangent at $(1,2)$ on circle, $x\left(1\right)-h(x+1)+h^2+2y-k(y+2)+k^2=r^2$

$=>y=\frac{(h-1)}{(2-k)}x+\frac{r^2-h^2-k^2+2h+4k}{(2-k)}......\left(4\right)$

Slope $m_1=\left(\frac{h-1}{2-k}\right).....\left(5\right)$

Tangent at $(1,2)$ on parabola, $2y=2x+2$

$=>y=x+1.........\left(6\right)$

Slope $m_2=1$

Now as circle cuts parabola orthogonaly $=>$angle between is$90°$,

$=>m_{1}m{}_2=-1$

$=>1\frac{(h-1)}{(2-k)}=-1$

$=>h-1-k+2=0$

$=>1+h-k=0$

Replace $h,k$ by $x,y$ for locus,

$=>1+x-y=0$

Now clearly from options,$(3,4)$ lies on the locus as,

$=>1+\left(3\right)-4=0$

Answer is $(3,4)$