Question

Find the equation of circle passing through (1, 1) belonging to the system of co-axial circles that are tangent at (2, 2) to the locus of the points of intersection of mutually perpendicular tangent to the circle $x^2 + y^2 = 4$.

Answer 1

The equation of the circle is: $x^2 + y^2 - 3x - 3y + 4 = 0$

Answer 2

1. The locus of points of intersection of mutually perpendicular tangents to a circle $x^2+y^2=r^2$ is its director circle, given by $x^2+y^2=2r^2$. For the given circle $x^2+y^2=4$ ($r^2=4$), the director circle is $x^2+y^2=2(4)=8$.
2. The tangent to the director circle $x^2+y^2=8$ at the point $(2,2)$ is found using the formula $xx_1+yy_1=r^2$. This gives $x(2)+y(2)=8$, which simplifies to $2x+2y=8$, or $x+y=4$. This line is the radical axis of the co-axial system of circles.
3. A system of co-axial circles tangent to a line $L=0$ at a point $(x_1, y_1)$ can be represented by the equation $(x-x_1)^2+(y-y_1)^2 + \lambda L = 0$. Here, $(x_1, y_1)=(2,2)$ and $L=x+y-4$. Thus, the equation of the co-axial system is $(x-2)^2+(y-2)^2 + \lambda(x+y-4) = 0$.
4. The required circle passes through the point $(1,1)$. Substituting these coordinates into the co-axial system's equation to find $\lambda$: $(1-2)^2 + (1-2)^2 + \lambda(1+1-4) = 0$ $(-1)^2 + (-1)^2 + \lambda(2-4) = 0$ $1 + 1 + \lambda(-2) = 0$ $2 - 2\lambda = 0 \implies \lambda = 1$
5. Substitute $\lambda=1$ back into the co-axial system equation: $(x-2)^2 + (y-2)^2 + 1(x+y-4) = 0$ Expanding and simplifying: $(x^2 - 4x + 4) + (y^2 - 4y + 4) + (x+y-4) = 0$ $x^2 + y^2 - 4x + x - 4y + y + 4 + 4 - 4 = 0$ $x^2 + y^2 - 3x - 3y + 4 = 0$