Q.A point moves such that the sum of the squares of its distances from the points $(1,2)$ and $(-1,-2)$ is always $26$. Find the equation of its locus.

Here both distances are squared and added, so the $x$ and $y$ terms will not cancel the way they do in a simple equidistant condition — expect a circle, since $x^2$ and $y^2$ keep equal coefficients.

Let $P(x,y)$. Condition: $(x-1)^2+(y-2)^2+(x+1)^2+(y+2)^2=26$. Expanding and adding gives $2x^2+2y^2+10=26$, so $x^2+y^2=8$.

Let $P(x,y)$. The given condition is:

Expand each squared term:

• $(x-1)^2 = x^2-2x+1$
• $(y-2)^2 = y^2-4y+4$
• $(x+1)^2 = x^2+2x+1$
• $(y+2)^2 = y^2+4y+4$

Adding all four: $2x^2 + 2y^2 + (-2x+2x) + (-4y+4y) + (1+4+1+4) = 2x^2+2y^2+10$.

So the condition becomes $2x^2+2y^2+10 = 26$, i.e. $2x^2+2y^2=16$, which simplifies to:

This is a circle centred at the origin with radius $\sqrt{8}=2\sqrt2$.

Independent check. Take a point that should satisfy $x^2+y^2=8$, e.g. $(2,2)$ (since $4+4=8$). Distance-squared from $(2,2)$ to $(1,2)$ is $(2-1)^2+(2-2)^2=1$. Distance-squared from $(2,2)$ to $(-1,-2)$ is $(2+1)^2+(2+2)^2=9+16=25$. Sum $=1+25=26$ ✓, exactly matching the given constant, which confirms the derived locus equation independently of the expansion above.

A frequent error is stopping after expanding only one or two of the four squared terms, or mis-adding the cross terms $-2x$ and $+2x$ (which must cancel) — always expand all four terms fully before combining, and verify the linear ($x$, $y$) terms cancel exactly, since a genuine circle centred at the origin has none.