JEE Main 2016. Previous Year Question.
Problem
Let $P$ be the point on the parabola $y^2=8x$ which is at a minimum distance from the centre $C$ of the circle $x^2+(y+6)^2=1$. Then the equation of the circle, passing through $C$ and having its centre at $P$, is:
Key insight. Rather than working with the distance formula directly (which involves a square root), parametrise the parabola and minimise the squared distance instead — a simple polynomial in the parameter. Since squaring is a monotonic operation for non-negative distances, wherever the squared distance is minimised, so is the actual distance.
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Approach
Identify the circle’s centre $C=(0,-6)$. Parametrise a point on the parabola $y^2=8x$ as $(2t^2,4t)$, and write the squared distance from this point to $C$ as a function of $t$. Differentiate, set the derivative to zero, and verify a minimum using the second derivative. Use the resulting point $P$ as the new circle’s centre, with $C$ giving the radius.
Solution
Step 1 — Identify the circle’s centre
For $x^2+(y+6)^2=1$: centre $C=(0,-6)$, radius $1$.
Step 2 — Parametrise the parabola
For $y^2=8x=4(2)x$, use $(at^2,2at)$ with $a=2$: a general point is $(2t^2,4t)$.
Step 3 — Write the squared distance to C
$$D^2 = (2t^2-0)^2+(4t-(-6))^2 = 4t^4+(4t+6)^2 = 4t^4+16t^2+48t+36$$
Step 4 — Differentiate and set to zero
$$\frac{d(D^2)}{dt} = 16t^3+32t+48 = 0$$
Dividing by 16:
$$t^3+2t+3=0$$
Step 5 — Find a root by inspection
Testing $t=-1$: $(-1)^3+2(-1)+3 = -1-2+3=0$ ✓
(The remaining quadratic factor has no real roots, so $t=-1$ is the only critical point.)
Step 6 — Verify it’s a minimum
$$\frac{d^2(D^2)}{dt^2} = 48t^2+32$$
At $t=-1$: $48+32=80>0$, confirming a minimum.
Step 7 — Find the point P
At $t=-1$: $P = (2(-1)^2,\ 4(-1)) = (2,-4)$
Step 8 — Find the radius of the new circle
The new circle has centre $P=(2,-4)$ and passes through $C=(0,-6)$:
$$r^2 = (2-0)^2+(-4-(-6))^2 = 4+4=8$$
Step 9 — Write the equation of the new circle
$$(x-2)^2+(y+4)^2 = 8$$
Expanding:
$$x^2-4x+4+y^2+8y+16 = 8$$
$$x^2+y^2-4x+8y+12=0$$
Answer
$$x^2+y^2-4x+8y+12=0$$
Common mistakes
- Minimising the actual distance $D$ instead of $D^2$. This is mathematically equivalent but requires differentiating a square root, making the algebra significantly messier — minimising $D^2$ avoids this entirely.
- Forgetting to verify the critical point is a minimum (not a maximum) using the second derivative. With a cubic equation having only one real root, it’s tempting to assume it must be the minimum, but explicitly checking prevents an error if the cubic had produced multiple real roots.
Practise next
- Let $Q$ be the point on the parabola $x^2=4y$ at minimum distance from the centre of the circle $(x-3)^2+y^2=1$. Find the equation of the circle centred at $Q$ that passes through that centre, using the same parametrise-and-minimise approach.
Show answer
$(x-2)^2+(y-1)^2=2$. Points of the parabola are $(2t,t^2)$, and the centre of the given circle is $(3,0)$.
Minimising $(2t-3)^2+t^4$ gives $4(2t-3)+4t^3=0$, that is $t^3+2t-3=0$. This factors as $(t-1)(t^2+t+3)=0$, and the quadratic has no real root, so $t=1$ is the only critical point — hence the minimum.
So $Q=(2,1)$, and the radius is the distance from $Q$ to $(3,0)$: $\sqrt{1+1}=\sqrt2$. The circle is $(x-2)^2+(y-1)^2=2$.
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