Condition for Orthogonal Intersection

When two circles intersect orthogonally, it means that at the points of intersection, the tangents to the circles are perpendicular to each other. This condition has important implications in geometry and can be described using algebraic equations.

Understanding Orthogonal Intersection

Two circles are said to intersect orthogonally if, at each point of intersection, the angle between their tangents is 90 degrees. This is equivalent to saying that the radius of each circle at the point of intersection is perpendicular to the radius of the other circle.

Mathematical Condition

For two circles with equations:

$$ (C_1): x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 $$ $$ (C_2): x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 $$

The condition for these circles to intersect orthogonally is given by:

$$ 2(g_1g_2 + f_1f_2) = c_1 + c_2 $$

This formula arises from the fact that the product of the slopes of the tangents at the points of intersection must be -1 (since they are perpendicular).

Table of Differences and Important Points

Feature	Orthogonal Intersection	General Intersection
Tangents	Perpendicular at the intersection points	Not necessarily perpendicular
Radii	Radii at the intersection points are perpendicular to each other	Radii are not necessarily perpendicular
Algebraic Condition	$2(g_1g_2 + f_1f_2) = c_1 + c_2$	No specific condition for general intersection
Slope Product	Product of slopes of tangents is -1	Product of slopes is not necessarily -1

Examples

Example 1: Verifying Orthogonal Intersection

Consider two circles given by:

$$ (C_1): x^2 + y^2 - 4x + 6y - 12 = 0 $$ $$ (C_2): x^2 + y^2 + 2x - 2y - 15 = 0 $$

To check if they intersect orthogonally, we can use the condition:

$$ 2(g_1g_2 + f_1f_2) = c_1 + c_2 $$

For $(C_1)$, we have $g_1 = -2$, $f_1 = 3$, and $c_1 = -12$. For $(C_2)$, we have $g_2 = 1$, $f_2 = -1$, and $c_2 = -15$.

Plugging these values into the condition:

$$ 2((-2)(1) + (3)(-1)) = -12 - 15 $$ $$ 2(-2 - 3) = -27 $$ $$ 2(-5) = -27 $$ $$ -10 \neq -27 $$

Since the left side does not equal the right side, the circles do not intersect orthogonally.

Example 2: Constructing Orthogonally Intersecting Circles

Suppose we want to construct a circle that intersects the circle $(C_1): x^2 + y^2 - 4x + 6y - 12 = 0$ orthogonally. We can choose parameters for the second circle $(C_2)$ such that the condition is satisfied.

Let's assume $(C_2)$ has the form $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ and we want to find $g_2$, $f_2$, and $c_2$.

Using the condition for orthogonal intersection:

$$ 2(g_1g_2 + f_1f_2) = c_1 + c_2 $$

We already know that for $(C_1)$, $g_1 = -2$, $f_1 = 3$, and $c_1 = -12$. Let's choose $g_2 = 2$ and $f_2 = -3$ to simplify our condition:

$$ 2((-2)(2) + (3)(-3)) = -12 + c_2 $$ $$ 2(-4 - 9) = -12 + c_2 $$ $$ 2(-13) = -12 + c_2 $$ $$ -26 = -12 + c_2 $$ $$ c_2 = -14 $$

Thus, a circle that intersects $(C_1)$ orthogonally could have the equation:

$$ (C_2): x^2 + y^2 + 4x - 6y - 14 = 0 $$

In summary, the condition for orthogonal intersection of two circles is a specific algebraic relationship between the coefficients of their equations. This condition ensures that the tangents to the circles at their points of intersection are perpendicular.