NIMCET 2023 — Mathematics PYQ

Step 1: Define the variables for the moving circle

Let the centre of the variable moving circle be $P(h,k)$ and its radius be $r$.

Since this variable circle touches the X-axis, the radius of the circle must be equal to the absolute value of its y-coordinate:

$$r=|k|$$

Assuming the circle lies above the X-axis to interact with the given circle centred at $(0,3)$, we have k > 0, so:

$$r=k$$

Step 2: Use the condition for touching circles

We are given a fixed circle with:

• Centre $C=(0,3)$

• Radius $r_1=2$

When two circles touch each other externally, the distance between their centres is equal to the sum of their radii:

$$PC=r+r_1$$

Using the distance formula to compute $PC$ between $P(h,k)$ and $C(0,3)$:

$$PC=\sqrt{(h-0{)}^2+(k-3{)}^2}$$

Substitute the values of $PC$, $r$, and $r_1$ into the condition:

$$\sqrt{h^2+(k-3{)}^2}=k+2$$

Step 3: Simplify the equation to find the locus

Square both sides of the equation to eliminate the radical sign:

$$h^2+(k-3{)}^2=(k+2{)}^2$$

Expand both quadratic expressions:

$$h^2+(k^2-6k+9)=k^2+4k+4$$

Cancel out $k^2$ from both sides:

$$h^2-6k+9=4k+4$$

Rearrange the terms to group $k$ together:

$$h^2=4k+6k+4-9$$

$$h^2=10k-5$$

To find the general equation of the locus, replace $(h,k)$ with standard coordinates $(x,y)$:

$$x^2=10y-5$$

Conclusion

The equation $x^2=10y-5$ represents a second-degree polynomial equation where one variable is squared and the other is linear. This is the standard form of a parabola opening upwards.

Correct Answer:

Correct Option: C