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(C) Let the base of the triangle be $BC$ of length $2c$ along the $x$-axis,with the midpoint at the origin $(0,0)$.Let the sum of the other two sides

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ellipse

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(C) Let the base of the triangle be $BC$ of length $2c$ along the $x$-axis,with the midpoint at the origin $(0,0)$.Let the sum of the other two sides $AB + AC = 2a$,where $a > c$.The vertices are $B(-c, 0)$ and $C(c, 0)$.The locus of the third vertex $A(x, y)$ is an ellipse given by $\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$.The coordinates of the incenter $I(h, k)$ are given by $\frac{ax_A + bx_B + cx_C}{a+b+c}$.For a triangle with a fixed base and a constant sum of sides,the incenter $I$ moves such that it satisfies the equation of an ellipse.Thus,the locus of the incenter is an ellipse.

Given the base of a triangle and the sum of its other two sides,the locus of the center of its incircle is:

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