The locus of the point from which mutually perpendicular tangents can be drawn to the circle2 2 4 6 3 0x y x y    i

To find the locus of a point from which mutually perpendicular tangents can be drawn to the given circle, let's follow the necessary steps.

1. Break Down the Problem

We are given the equation of a circle and need to determine the locus of points that can draw mutually perpendicular tangents to this circle.

2. Relevant Concepts

The general equation of a circle is given by: $(x - h)^2 + (y - k)^2 = r^2$ where $(h, k)$ is the center and $r$ is the radius.

For mutually perpendicular tangents drawn from a point $(x_1, y_1)$ to a circle, the condition is related to the angle formed by the radius at the point of tangency.

3. Analysis and Detail

First, rewrite the given circle equation: $2x^2 + 2y^2 - 4x - 6y + 3 = 0$ Dividing through by 2, we have: $x^2 + y^2 - 2x - 3y + \frac{3}{2} = 0$ Now, let's complete the square:

For $x$: $x^2 - 2x \Rightarrow (x - 1)^2 - 1$

For $y$: $y^2 - 3y \Rightarrow (y - \frac{3}{2})^2 - \frac{9}{4}$

Substituting back, we obtain: $(x - 1)^2 - 1 + (y - \frac{3}{2})^2 - \frac{9}{4} + \frac{3}{2} = 0$ Simplifying this gives: $(x - 1)^2 + (y - \frac{3}{2})^2 = \frac{9}{4} - \frac{3}{4} = \frac{6}{4} = \frac{3}{2}$

Thus, the center $(h, k)$ of the circle is $(1, \frac{3}{2})$ with a radius $r = \sqrt{\frac{3}{2}}$.

4. Verify and Summarize

For a point $(x_1, y_1)$ to have mutually perpendicular tangents to the circle, the following condition should hold: $(x_1 - 1)^2 + \left(y_1 - \frac{3}{2}\right)^2 = r^2 + (r^2) = 2r^2$

Substituting the computed value of $r^2 = \frac{3}{2}$: $(x_1 - 1)^2 + \left(y_1 - \frac{3}{2}\right)^2 = 2 \times \frac{3}{2} = 3$

Final Answer

The locus of the points from which mutually perpendicular tangents can be drawn to the circle is given by the equation: $(x - 1)^2 + \left(y - \frac{3}{2}\right)^2 = 3$