Animated Solution for Mathematics - Circles: If a circle passes through the point $(a,b)$ and cuts the circle $x^2+y^2=k^2$ orthogonally, then the equation of the locus of its centre is

To begin, we define the variable circle with the general equation:

The center of this circle is $C(-g,-f)$. Our objective is to determine the path traced by this center, which requires establishing a relationship between the parameters $g$, $f$, and the constants $a$, $b$, and $k$.

Two circles intersect orthogonally if the square of the distance between their centers equals the sum of the squares of their radii. Algebraically, for two circles $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ and $x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$, the condition is:

Our fixed circle is $x^2+y^2-k^2=0$. Comparing this to the general form, we identify $g_2=0$, $f_2=0$, and $c_2=-k^2$.

Substituting these values into the orthogonality condition yields:

This simplifies to the elegant result:

The variable circle must pass through the fixed point $P(a,b)$. Consequently, the coordinates of $P$ must satisfy the equation of the circle:

By substituting our previously derived value $c=k^2$ into this equation, we obtain:

We define the locus of the center $(-g,-f)$ as $(x,y)$. This implies the following substitutions:

Substituting these into the anchor point equation, we get:

Simplifying the expression leads to the final equation of the locus:

The resulting locus is a straight line. This confirms that the center of the variable circle moves along a linear path defined by the coordinates of $P$ and the radius of the fixed circle.