IPMAT Indore 2023 · Quantitative Ability · Question 34

Step 1 — complete the square in both variables.

\left({x}^{2}-2x \right)+\left({y}^{2}-4y \right)+5=0

{\left( x-1 \right)}^{2}-1+{\left( y-2 \right)}^{2}-4+5=0

{\left( x-1 \right)}^{2}+{\left( y-2 \right)}^{2}=0

Step 2 — interpret. The right-hand side is zero, so the “radius” is zero. A sum of two squares equals zero only when each is zero:

x=1, y=2

The equation is satisfied by the single point \left( 1,2 \right) and nothing else — a point circle, the degenerate case.

Step 3 — why the others are wrong.

(a) a circle — would need {r}^{2}>0. Here r=0, so there is no curve, only a point.

(b) an ellipse — would need different coefficients on {x}^{2} and {y}^{2}; here both are 1.

(d) a pair of straight lines — a real pair needs the expression to factor into two real linear factors; here the only real solution is one point.

Answer: Option 3 — a point

Shortcut. Compute {g}^{2}+{f}^{2}-c before anything else. For {x}^{2}+{y}^{2}+2gx+2fy+c=0 it is the square of the radius:

{g}^{2}+{f}^{2}-c=1+4-5=0

>0 → a real circle

=0 → a point circle ✔

<0 → imaginary, no real locus

Here it is exactly 0, so the answer is a point — one subtraction’s worth of work.

This degenerate-conic question appears regularly, and the giveaway is always a constant term that makes the expression collapse. Whenever a circle question offers “a point” among the options, check {g}^{2}+{f}^{2}-c first; the setter is usually testing precisely that.