Q.A point $P$ moves so that its distance from the fixed point $C(1, 2)$ is always $5$ units. Find the equation of its locus.

A point at a fixed distance from a fixed point traces a circle; take $P(x,y)$, set the distance $PC = 5$, and square.

$(x-1)^2 + (y-2)^2 = 25$, which expands to $x^2 + y^2 - 2x - 4y - 20 = 0$.

Let $P(x, y)$ be any point on the locus. The given condition is $PC = 5$, where $C(1,2)$ is fixed. Squaring, $PC^2 = 25$:

This is already the equation of the locus in the neat centre–radius form of a circle with centre $(1,2)$ and radius $5$. Expanding to the general form:

Check (independent). Take a point that should be exactly $5$ units from $C(1,2)$, e.g. $(6, 2)$ (which is $5$ to the right of $C$). Substituting: $36 + 4 - 12 - 8 - 20 = 0$. It satisfies the equation, confirming the locus.

Squaring $5$ incorrectly (using $25$ is right; some write $10$), or making an arithmetic slip when combining the constants $1 + 4 - 25 = -20$. Leaving the answer in the centre–radius form $(x-1)^2+(y-2)^2=25$ is also perfectly acceptable unless the general form is specifically asked.